{"id":2816,"date":"2022-11-24T10:00:43","date_gmt":"2022-11-24T04:30:43","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2816"},"modified":"2022-11-25T16:47:27","modified_gmt":"2022-11-25T11:17:27","slug":"types-of-quadrilaterals","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/types-of-quadrilaterals\/","title":{"rendered":"How Many Types Of Quadrilaterals Are There"},"content":{"rendered":"<h2><strong>How Many Types Of <a href=\"https:\/\/www.aplustopper.com\/what-is-a-quadrilateral\/\">Quadrilaterals<\/a> Are There<\/strong><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-12007 aligncenter\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/10\/what-is-a-quadrilateral-2.jpg\" alt=\"\" width=\"650\" height=\"397\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/10\/what-is-a-quadrilateral-2.jpg 650w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/10\/what-is-a-quadrilateral-2-300x183.jpg 300w\" sizes=\"(max-width: 650px) 100vw, 650px\" \/><\/p>\n<ul>\n<li>A quadrilateral is a figure bounded by four line segments such that no three of them are parallel.<\/li>\n<li>Two sides of quadrilateral are consecutive or adjacent sides, if they have a common point (vertex).<\/li>\n<li>Two sides of a quadrilateral are opposite sides, if they have no common end-point (vertex).<\/li>\n<li>The consecutive angles of a quadrilateral are two angles which include a side in their intersection.<\/li>\n<li>In other words, two angles are consecutive, if they have a common arm.<\/li>\n<li>Two angles of a quadrilateral are said to be opposite angles if they do not have a common arm.<\/li>\n<li>The sum of the four angles of a quadrilateral is 360\u00ba.<\/li>\n<li>A quadrilateral having exactly one pair of parallel sides, is called a trapezium.<\/li>\n<li>A trapezium is said to be an isoscels trapezium, if its non-parallel sides are equal.<\/li>\n<li>A quadrilateral is a parallelogram if its both pairs of opposite sides are parallel.<\/li>\n<li>A parallelogram having all sides equal is called a rhombus.<\/li>\n<li>A parallelogram whose each angle is a right angle, is called a rectangle.<\/li>\n<li>A square is a rectangle with a pair of adjacent sides equal.<\/li>\n<li>A quadrilateral is a kite if it has two pairs of equal adjacent sides and unequal opposite sides.<\/li>\n<li>A diagonal of a parallelogram divides it into two congruent triangles.<\/li>\n<li>In a parallelogram, opposite sides are equal.<\/li>\n<li>The opposite angles of a parallelogram are equal.<\/li>\n<li>The diagonals of a parallelogram bisect each other.<\/li>\n<li>In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.<\/li>\n<li>If diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle.<\/li>\n<li>The angle bisectors of a parallegram form a rectangle.<\/li>\n<li>A quadrilateral is a parallelogam if its opposite sides are equal.<\/li>\n<li>A quadrilateral is a parallelogram if its opposite angles are equal.<\/li>\n<li>If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.<\/li>\n<li>A quadrilateral is a parallelogram, if its one pair of opposite sides are equal and parallel.<\/li>\n<li>Each of the four angles of a rectangel is a right angle.<\/li>\n<li>Each of the four sides of a rhombus is of the same length.<\/li>\n<li>Each of the angles of a square is a right angle and each of the four sides is of the same length.<\/li>\n<li>The diagonals of a rectangle are of equal length.<\/li>\n<li>If the two diagonals of parallelogram are equal, it is a rectangle.<\/li>\n<li>The diagonals of a rhombus are perpendicular to each other.<\/li>\n<li>If the diagonals of a parallelogram are perpendicular, then it is a rhombus.<\/li>\n<li>The diagonals of a square are equal and perpendicular to each other.<\/li>\n<li>If the diagonals of a parallelogram are equal and intersect at right angles then the parallelogram is a square.<\/li>\n<li>A diagonal of a parallelogram divides it into two triangles of equal area.<\/li>\n<li>For each base of a parallelogram, the corresponding altitude is the line segment from a point on the base, perpendicular to the line containing the opposite side.<\/li>\n<li>Parallelograms on the same base and between the same parallels are equal in area.<\/li>\n<li>A parallelogram and a rectangle on the same base and between the same parallels are equal in area.<\/li>\n<li>The area of a parallelogram is the product of its base and the corresponding altitude.<\/li>\n<li>Parallelograms on equal bases and between the same parallels are equal in area.<\/li>\n<\/ul>\n<p><strong>Read More:\u00a0<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/www.aplustopper.com\/different-types-of-quadrilaterals\/\">Different Kinds of Quadrilateral<\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/properties-of-cyclic-quadrilaterals\/\">Properties of Cyclic Quadrilaterals<\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/lesson\/rs-aggarwal-class-9-solutions-quadrilaterals-and-parallelograms\/\">RS Aggarwal Class 9 Solutions Quadrilaterals and Parallelograms<\/a><\/li>\n<\/ul>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/yiREqzDsMP8?feature=oembed\" frameborder=\"0\" allow=\"autoplay; encrypted-media\" allowfullscreen><\/iframe><\/p>\n<h2><strong>Types Of Quadrilaterals Example Problems With Solutions<\/strong><\/h2>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 2 : 4 : 5 : 7. Find the measure of each angles of the quadrilateral.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0We have \u2220A : \u2220B : \u2220C : \u2220D = 2 : 4 : 5 : 7.<br \/>\nSo, let \u2220A = 2x\u00ba, \u2220B = 4x\u00ba, \u2220C = 5x\u00ba,\u00a0\u2220D = 7x\u00ba.<br \/>\n\u2234 \u2220A + \u2220B + \u2220C + \u2220D = 360\u00ba<br \/>\n\u21d2 x + 4x + 5x + 7x = 360\u00ba<br \/>\n\u21d2 18x = 360\u00ba<br \/>\n\u21d2 x = 20\u00ba<br \/>\nThus, the angles are:<br \/>\n\u2220A = 40\u00ba, \u2220B = (4 \u00d7 20)\u00ba = 80\u00ba,<br \/>\n\u2220C = (5 \u00d7 20)\u00ba = 100\u00ba<br \/>\nand, \u2220D = (7x)\u00ba = (7 \u00d7 20)\u00ba = 140\u00ba<\/p>\n<p><strong>Example 2: \u00a0 \u00a0<\/strong>The sides BA and DC of a quadrilateral ABCD are produced as shown in fig.<br \/>\nProve that a + b = x + y.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Join BD. In \u2206ABD, we have<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8054\/29633522156_0df46a3485_o.png\" alt=\"Types-Of-Quadrilaterals-Example-2\" width=\"247\" height=\"185\" \/><br \/>\n\u2220ABD + \u2220ADB = b\u00ba \u00a0 \u00a0 \u00a0 &#8230;.(i)<br \/>\nIn \u2206CBD, we have<br \/>\n\u2220CBD + \u2220CDB = a\u00ba \u00a0 \u00a0 \u00a0 &#8230;.(ii)<br \/>\nAdding (i) and (ii), we get<br \/>\n(\u2220ABD + \u2220CBD) + (\u2220ADB + \u2220CDB) = a\u00ba + b\u00ba<br \/>\n\u21d2 x\u00ba + y\u00ba = a\u00ba + b\u00ba<br \/>\nHence, x + y = a + b<\/p>\n<p><strong>Example 3: \u00a0 \u00a0<\/strong>In a quadrilateral ABCD, AO and BO are the bisectors of \u2220A and \u2220B respectively. Prove that \u2220AOB = 1\/2 (\u2220C + \u2220D).<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0In \u2206AOB, we have<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8436\/29042743384_0e9c24b546_o.png\" alt=\"Types-Of-Quadrilaterals-Example-3\" width=\"191\" height=\"179\" \/><br \/>\n\u2220AOB + \u22201 + \u22202 = 180\u00ba<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2823 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-3-1.png\" alt=\"types-of-quadrilaterals-example-3-1\" width=\"348\" height=\"506\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-3-1.png 348w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-3-1-206x300.png 206w\" sizes=\"(max-width: 348px) 100vw, 348px\" \/><\/p>\n<p><strong>Example 4: \u00a0 \u00a0<\/strong>In figure bisectors of \u2220B and \u2220D of quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that<br \/>\n\u2220P + \u2220Q = 1\/2 (\u2220ABC + \u2220ADC)<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8438\/29587870571_1fd7159d6b_o.png\" alt=\"Types-Of-Quadrilaterals-Example-4\" width=\"204\" height=\"168\" \/><br \/>\n<strong>Solution:<\/strong><br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8540\/29044988133_a7064e59fc_o.png\" alt=\"Types-Of-Quadrilaterals-Example-4-1\" width=\"364\" height=\"473\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8120\/29634068036_2af84fa0dc_o.png\" alt=\"Types-Of-Quadrilaterals-Example-4-2\" width=\"371\" height=\"179\" \/><\/p>\n<p><strong>Example 5: \u00a0 \u00a0<\/strong>In a parallelogram ABCD, prove that sum of any two consecutive angles is 180\u00ba.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Since ABCD is a parallelogram. Therefore,<br \/>\nAD || BC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8141\/29042743134_680892456d_o.png\" alt=\"Types-Of-Quadrilaterals-Example-5\" width=\"209\" height=\"191\" \/><br \/>\nNow, AD || BC and transversal AB intersects them at A and B respectively.<br \/>\n\u2234 \u2220A + \u2220B = 180\u00ba<br \/>\n[\u2235 Sum of the interior angles on the same side of the transversal is 180\u00ba]<br \/>\nSimilarly, we can prove that<br \/>\n\u2220B + \u2220C = 180\u00ba, \u2220C + \u2220D = 180\u00ba and<br \/>\n\u2220D + \u2220A = 180\u00ba.<\/p>\n<p><strong>Example 6: \u00a0 \u00a0<\/strong>In a parallelogram ABCD, \u2220D = 115\u00ba, determine the measure of \u2220A and \u2220B.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Since the sum of any two consecutive angles of a parallelogram is 180\u00ba. Therefore,<br \/>\n\u2220A + \u2220D = 180\u00ba and \u2220A + \u2220B = 180\u00ba<br \/>\nNow, \u2220A + \u2220D = 180\u00ba<br \/>\n\u21d2 \u2220A + 115\u00ba = 180\u00ba \u00a0 \u00a0 \u00a0[\u2235 \u2220D = 115\u00ba (given)]<br \/>\n\u21d2 \u2220A = 65\u00ba and \u2220A + \u2220B = 180\u00ba<br \/>\n\u21d2 65\u00ba + \u2220B = 180\u00ba \u21d2 \u2220B = 115\u00ba<br \/>\nThus, \u2220A = 65\u00ba and \u2220B = 115\u00ba<\/p>\n<p><strong>Example 7: \u00a0 \u00a0<\/strong>In figure, AB = AC, \u2220EAD = \u2220CAD and<br \/>\nCD || AB. Show that ABCD is a parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8338\/29633521846_2d5602ca6b_o.png\" alt=\"Types-Of-Quadrilaterals-Example-7\" width=\"216\" height=\"224\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0In \u2206ABC, AB = AC [Given]<br \/>\n\u21d2 \u2220ABC = \u2220ACB \u00a0 \u00a0 \u00a0 \u00a0\u2026.(1)<br \/>\n(Angles opposite the equal sides are equal)<br \/>\n\u2220EAD = \u2220CAD[Given] \u00a0 \u00a0 \u00a0 \u00a0 \u2026.(2)<br \/>\nNow, \u2220EAC = \u2220ABC + \u2220ACB<br \/>\n[An exterior angle is equal to sum of two interior opposite angles of a triangles]<br \/>\n\u21d2 \u2220EAD + \u2220CAD = \u2220ABC + \u2220ACB<br \/>\n\u21d2 \u2220CAD + \u2220CAD = \u2220ACB + \u2220ACB<br \/>\nBy (1) and (2)<br \/>\n\u21d2 2\u2220CAD = 2\u2220ACB<br \/>\n\u21d2 \u2220CAD = \u2220ACB<br \/>\n\u21d2 BC | | AD<br \/>\nAlso, CD | | AB \u00a0 \u00a0 \u00a0 \u00a0 [Given]<br \/>\nThus, we have both pairs of opposite sides of quadrilateral ABCD parallel. Therefore, ABCD is a parallelogram.<\/p>\n<p><strong>Example 8: \u00a0 \u00a0<\/strong>ABCD is a parallelogram and line segments AX,CY are angle bisector of \u2220A and \u2220C respectively then show AX || CY.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8519\/29633521786_26358718e6_o.png\" alt=\"Types-Of-Quadrilaterals-Example-8\" width=\"253\" height=\"197\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Since opposite angles are equal in a parallelogram. Therefore, in parallelogram ABCD, we have \u2220A = \u2220C<br \/>\n\u21d2 1\/2 \u2220A = 1\/2 \u2220C<br \/>\n\u21d2 \u22201 = \u22202 \u00a0 \u00a0 &#8230;.(i)<br \/>\n[\u2235 AX and CY are bisectors of \u2220A and \u2220C respectively]<br \/>\nNow, AB || DC and the transversal CY intersects them.<br \/>\n\u2234 \u22202 = \u22203 \u00a0 \u00a0\u00a0&#8230;(ii)<br \/>\n[\u2235 Alternate interior angles are equal]<br \/>\nFrom (i) and (ii), we get<br \/>\n\u22201 = \u22203<br \/>\nThus, transversal AB intersects AX and YC at A and Y such that \u22201 = \u22203 i.e. corresponding angles are equal.<br \/>\n\u2234 AX || CY<\/p>\n<p><strong>Example 9: \u00a0 \u00a0<\/strong>In the adjoining figure, a point O is taken inside an equilateral quad. ABCD such that OB = OD. Show that A, O and C are in the same straight line.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8156\/29042742834_824e43e98e_o.png\" alt=\"Types-Of-Quadrilaterals-Example-9\" width=\"205\" height=\"182\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given a quad. ABCD in which AB = BC<br \/>\n= CD = DA and O is a point within it such that OB = OD.<br \/>\nTo prove \u2220AOB + \u2220COB = 180\u00ba<br \/>\nProof In \u2206OAB and OAD, we have<br \/>\nAB = AD \u00a0 \u00a0 (given)<br \/>\nOA = OA\u00a0 \u00a0(common)<br \/>\nand OB = OD (given)<br \/>\n\u2234 \u2206OAB \u2245 \u2206OAD<br \/>\n\u2234 \u2220AOB = \u2220AOD \u00a0 \u00a0 \u00a0 \u00a0\u2026.(i) \u00a0 \u00a0 (c.p.c.t.)<br \/>\nSimilarly, \u2206OBC \u2245 \u2206ODC<br \/>\n\u2234\u2220COB = \u2220COD \u00a0 \u00a0 \u00a0 \u00a0 \u2026.(ii)<br \/>\nNow, \u2220AOB + \u2220COB + \u2220COD + \u2220AOD<br \/>\n= 360\u00ba \u00a0 \u00a0 \u00a0[\u2220 at a point]<br \/>\n\u21d2 2(\u2220AOB + \u2220COB) = 360\u00ba<br \/>\n\u21d2 \u2220AOB + \u2220COB = 180\u00ba<\/p>\n<p><strong>Example 10: \u00a0 \u00a0<\/strong>In figure AN and CP are perpendiculars to the diagonal BD of a parallelogram ABCD. Prove that:<br \/>\n(i) \u2206ADN \u2245 \u2206CBP (ii) AN = CP<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8580\/29378303710_6b0a4fb4f2_o.png\" alt=\"Types-Of-Quadrilaterals-Example-10\" width=\"207\" height=\"151\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Since ABCD is a parallelogram.<br \/>\n\u2234 AD || BC<br \/>\nNow, AD || BC and transversal BD intersects them at B and D.<br \/>\n\u2234 \u22201 = \u22202<br \/>\n[\u2235 Alternate interior angles are equal]<br \/>\nNow, in \u2206s ADN and CBP, we have<br \/>\n\u22201 = \u22202<br \/>\n\u2220AND = \u2220CPD and, AD = BC<br \/>\n[\u2235 Opposite sides of a ||gm are equal]<br \/>\nSo, by AAS criterion of congruence<br \/>\n\u2206ADN \u2245 \u2206CBP<br \/>\nAN = CP<br \/>\n[\u2235 Corresponding parts of congruent triangles are equal]<\/p>\n<p><strong>Example 11: \u00a0 \u00a0<\/strong>In figure, ABCD is a trapezium such that\u00a0AB || CD and AD = BC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8072\/29633521646_b6802851cf_o.png\" alt=\"Types-Of-Quadrilaterals-Example-11\" width=\"250\" height=\"165\" \/><br \/>\nBE || AD and BE meets BC at E.<br \/>\nShow that (i) ABED is a parallelogram.<br \/>\n(ii) \u2220A + \u2220C = \u2220B + \u2220D = 180\u00ba.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Here, AB || CD (Given)<br \/>\n\u21d2 AB || DE \u00a0 \u00a0 \u00a0 \u2026.(1)<br \/>\nAlso, BE || AD (Given) \u00a0 \u00a0 \u2026.(2)<br \/>\nFrom (1) and (2),<br \/>\nABED is a parallelogram<br \/>\n\u21d2 AD = BE \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(3)<br \/>\nAlso, AD = BC (Given) \u00a0 \u00a0 \u00a0 \u2026.(4)<br \/>\nFrom (3) and (4),<br \/>\nBE = BC<br \/>\n\u21d2 \u2220BEC = \u2220BCE \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(5)<br \/>\nAlso, \u2220BAD = \u2220BED<br \/>\n(opposite angles of parallelogram ABED)<br \/>\ni.e., \u2220BED = \u2220BAD \u00a0 \u00a0 \u00a0 \u00a0\u2026.(6)<br \/>\nNow, \u2220BED + \u2220BEC = 180\u00ba (Linear pair of angles)<br \/>\n\u21d2 \u2220BAD + \u2220BCE = 180\u00ba<br \/>\nBy (5) and (6)<br \/>\n\u21d2 \u2220A + \u2220C = 180\u00ba<br \/>\nSimilarly, \u2220B + \u2220D = 180\u00ba<\/p>\n<p><strong>Example 12: \u00a0 \u00a0<\/strong>In figure ABCD is a parallelogram and\u00a0\u2220DAB = 60\u00ba. If the bisectors AP and BP of angles A and B respectively, meet at P on CD, prove that P is the mid-point of CD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c2.staticflickr.com\/8\/7774\/29558127232_a79668292c_o.png\" alt=\"Types-Of-Quadrilaterals-Example-12\" width=\"220\" height=\"156\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0We have, \u2220DAB = 60\u00ba<br \/>\n\u2220A + \u2220B = 180\u00ba<br \/>\n\u2234 60\u00ba + \u2220B = 180\u00ba \u21d2 \u2220B = 120\u00ba<br \/>\nNow, AB || DC and transversal AP intersects them.<br \/>\n\u2234 \u2220PAB = \u2220APD<br \/>\n\u21d2 \u2220APD = 30\u00ba \u00a0 \u00a0 \u00a0[\u2235 \u2220PAB = 30\u00ba]<br \/>\nThus, in \u2206APD, we have<br \/>\n\u2220PAD = \u2220APD \u00a0 \u00a0 \u00a0[Each equal to 30\u00ba]<br \/>\n\u21d2 AD = PD \u00a0 \u00a0 \u00a0&#8230;. (i)<br \/>\n[\u2235 Angles opposite to equal sides are equal] Since BP is the bisector of \u2220B. Therefore,<br \/>\n\u2220ABP = \u2220PBC = 60\u00ba<br \/>\nNow, AB || DC and transversal BP intersects them.<br \/>\n\u2234 \u2220CPB = \u2220ABP<br \/>\n\u21d2 \u2220CPB = 60\u00ba \u00a0 \u00a0 \u00a0[\u2235 \u2220ABP = 60\u00ba]<br \/>\nThus, in \u2206CBP, we have<br \/>\n\u2220CBP = \u2220CPB \u00a0 \u00a0 \u00a0\u00a0[Each equal to 60\u00ba]<br \/>\n\u21d2 CP = BC<br \/>\n\u2235 [Sides opp, to equal angles are equal]<br \/>\n\u21d2 CP = AD &#8230;. (ii)<br \/>\n[\u2235 ABCD is a parallelogram\u00a0\u2234 AD = BC]<br \/>\nFrom (i) and (ii), we get<br \/>\nPD = CP<br \/>\n\u21d2 P is the mid point of CD.<\/p>\n<p><strong>Example 13: \u00a0 \u00a0<\/strong>Prove that the line segments joining the mid-point of the sides of a quadrilateral forms a parallelogram.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Points E, F, G and H are the mid-points of the sides AB, BC, CD and DA respectively, of the quadrilateral ABCD. We have to prove that EFGH is a parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8280\/29378303470_b31babba99_o.png\" alt=\"Types-Of-Quadrilaterals-Example-13\" width=\"287\" height=\"235\" \/><br \/>\nJoin the diagonal AC of the quadrilateral ABCD.<br \/>\nNow, in \u2206ABC, we have E and F mid-points of the sides BA and BC.<br \/>\n\u21d2 EF || AC<br \/>\nand EF = 1\/2 AC \u00a0 \u00a0 \u00a0&#8230;. (1)<br \/>\nSimilarly, from \u2206ADC, we have<br \/>\nGH || AC<br \/>\nand GH = 1\/2 AC \u00a0 \u00a0 \u00a0\u2026.(2)<br \/>\nThen from (1) and (2), we have<br \/>\nEF || GH<br \/>\nand EF = GH<br \/>\nThis proves that EFGH is a parallelogram.<\/p>\n<p><strong>Example 14: \u00a0 \u00a0<\/strong>In figure ABCD is a parallelogram and X, Y are the mid-points of sides AB and DC respectively. Show that AXCY is a parallelogram.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Since X and Y are the mid-points of AB and DC respectively. Therefore,<br \/>\nAX = 1\/2 AB and CY = 1\/2 DC \u00a0 \u00a0&#8230; (i)<br \/>\nBut, AB = DC \u00a0 \u00a0[\u2235 ABCD is a parallelogram]<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8006\/29558127202_e571889f42_o.png\" alt=\"Types-Of-Quadrilaterals-Example-14\" width=\"246\" height=\"157\" \/><br \/>\n\u21d2 1\/2 AB = 1\/2 DC<br \/>\n\u21d2 AX = CY \u00a0 \u00a0 \u00a0 \u00a0&#8230;. (ii)<br \/>\nAlso, AB || DC<br \/>\n\u21d2 AX || YC \u00a0 \u00a0 \u00a0 \u00a0&#8230;. (iii)<br \/>\nThus, in quadrilateral AXCY, we have<br \/>\nAX || YC and AX = YC<br \/>\n[From (ii) and (iii)]<br \/>\nHence, quadrilateral AXCY is a parallelogram.<\/p>\n<p><strong>Example 15: \u00a0 \u00a0<\/strong>Prove that the line segments joining the mid-points of the sides of a rectangle forms a rhombus.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0P, Q, R and S are the mid-points of the sides AB, BC, CD and DA of the rectangle ABCD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8513\/29558127172_bcc2278137_o.png\" alt=\"Types-Of-Quadrilaterals-Example-15\" width=\"270\" height=\"197\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8445\/29670035435_3e62e4db9a_o.png\" alt=\"Types-Of-Quadrilaterals-Example-15-1\" width=\"369\" height=\"391\" \/><\/p>\n<p><strong>Example 16: \u00a0 \u00a0<\/strong>In figure ABCD is a parallelogram and X and Y are points on the diagonal BD such that<br \/>\nDX = BY. Prove that<br \/>\n(i) AXCY is a parallelogram<br \/>\n(ii) AX = CY, AY = CX<br \/>\n(iii) \u2206AYB \u2245 \u2206CXD<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given : ABCD is a parallelogram. X and Y are points on the diagonal BD such that<br \/>\nDX = BY<br \/>\n<strong>To Prove:<\/strong><br \/>\n(i) AXCY is a parallelogram<br \/>\n(ii) AX = CY, AY = CX<br \/>\n(iii) \u2206AYB \u2245 \u2206CXD<br \/>\nC Construction : join AC to meet BD at O.<br \/>\n<strong>Proof:<\/strong><br \/>\n<strong>(i)<\/strong> We know that the diagonals of a parallelogram bisect each other. Therefore, AC and BD bisect each other at O.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8033\/29042742084_93daa785a7_o.png\" alt=\"Types-Of-Quadrilaterals-Example-16\" width=\"229\" height=\"150\" \/><br \/>\n\u2234 OB = OD<br \/>\nBut, BY = DX<br \/>\n\u2234 OB \u2013 BY = OD \u2013 DX<br \/>\n\u21d2 OY = OX<br \/>\nThus, in quadrilateral AXCY diagonals AC and XY are such that OX = OY and OA = OC i.e. the diagonals AC and XY bisect each other.<br \/>\nHence, AXCY is a parallelogram.<br \/>\n<strong>(ii)<\/strong> Since AXCY is a parallelogram<br \/>\n\u2234 AX = CY and AY = CX<br \/>\n<strong>(iii)<\/strong> In triangles AYB and CXD, we have<br \/>\nAY = CX [From (ii)]<br \/>\nAB = CD<br \/>\n[\u2235 ABCD is a parallelogram]<br \/>\nBY = DX [Given]<br \/>\nSo, by SSS-criterion of congruence, we have<br \/>\n\u2206AYB \u2245 \u2206CXD<\/p>\n<p><strong>Example 17: \u00a0 \u00a0<\/strong>In fig. ABC is an isosceles triangle in which\u00a0AB = AC. CP || AB and AP is the bisector<br \/>\nof exterior \u2220CAD of \u2206ABC. Prove that<br \/>\n\u2220PAC = \u2220BCA and ABCP is a parallelogram.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given : An isosceles \u2206ABC having<br \/>\nAB = AC.AP is the bisector of ext \u2220CAD and CP || AB.<br \/>\n<strong>To Prove:<\/strong> \u2220PAC = \u2220BCA and ABCP<br \/>\n<strong>Proof:<\/strong> In \u2206ABC, we have<br \/>\nAB = AC [Given]<br \/>\n\u21d2 \u22201 = \u22202 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;. (i)<br \/>\n\u2235 Angles opposite to equal sides in a \u2206 are equal<br \/>\nNow, in \u2206 ABC, we have<br \/>\next \u2220CAD = \u22201 + \u22202<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8546\/29378303170_1f2db6a303_o.png\" alt=\"Types-Of-Quadrilaterals-Example-17\" width=\"265\" height=\"217\" \/><br \/>\n\u21d2 ext \u2220CAD = 2\u22202 [\u2235 \u22201 = \u22202 (from (i))]<br \/>\n\u21d2 2\u22203 = 2\u22202<br \/>\n[\u2235 AP is the bisector of ext.\u2220CAD \u2234\u2220CAD = 2\u22203]<br \/>\n\u21d2 \u22203 = \u22202<br \/>\nThus, AC intersects lines AP and BC at A and C respectively such that \u22203 = \u22202<br \/>\ni.e., alternate interior angles are equal.<br \/>\nTherefore, \u00a0AP || BC.<br \/>\nBut, CP || AB \u00a0 \u00a0 \u00a0 \u00a0 [Gvien]<br \/>\nThus, ABCP is a quadrilateral such that<br \/>\nAP || BC and CP || AB.<br \/>\nHence, ABCP is a parallelogram.<\/p>\n<p><strong>Example 18: \u00a0 \u00a0<\/strong>In the given figure, ABCD is a square and \u2220PQR = 90\u00ba. If PB = QC = DR, prove that<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8748\/29378303040_04831ecb91_o.png\" alt=\"Types-Of-Quadrilaterals-Example-18\" width=\"201\" height=\"207\" \/><br \/>\n(i) QB = RC, \u00a0 \u00a0 (ii) PQ = QR, \u00a0 \u00a0 \u00a0 \u00a0(iii) \u2220QPR = 45\u00ba.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0BC = DC, CQ = DR \u21d2 BC \u2013 CQ = \u2206CDR<br \/>\n\u21d2 QB = RC<br \/>\nFrom \u2206CQR, \u2220RQB = \u2220QCR + \u2220QRC<br \/>\n\u21d2 \u2220RQP + \u2220PQB = 90\u00ba + \u2220QRC<br \/>\n\u21d2 90\u00ba + \u2220PQB = 90\u00ba + \u2220QRC<br \/>\nNow, \u2206RCQ \u2245\u00a0\u2206QBP and therefore,<br \/>\nQR = PQ<br \/>\nPQ = QR \u21d2 \u2220QPR = \u2220PRQ<br \/>\nBur, \u2220QPR + \u2220PRQ = 90\u00ba.<br \/>\nSo, \u2220QPR = 45\u00ba<\/p>\n<p><strong>Example 19: \u00a0 \u00a0<\/strong>Prove that in a parallelogram<br \/>\n(i) opposite sides are equal<br \/>\n(ii) opposite angles are equal<br \/>\n(iii) each diagonal bisects the parallelogram<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>Given:<\/strong> A ||gm ABCD in which AB || DC and AD || BC.<br \/>\n<strong>To prove:<\/strong> (i) AB = CD and BC = AD;<br \/>\n(ii) \u2220B = \u2220D and \u2220A = \u2220C,<br \/>\n(iii) \u2206ABC = \u2206CDA and \u2206ABD = \u2206CDB<br \/>\nConstruction join A and C.<br \/>\nIn \u2206ABC and CDA, we have,<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8063\/29558127022_7885ab5013_o.png\" alt=\"Types-Of-Quadrilaterals-Example-19\" width=\"309\" height=\"167\" \/><br \/>\n\u22201 = \u22202<br \/>\n[Alt. int. \u2220, as AB || DC and CA cuts them]<br \/>\n\u22203 = \u22204<br \/>\n[Alt. int. \u2220, as BC || AD and CA cuts them]<br \/>\nAC = CA (common)<br \/>\n\u2234 \u2206ABC \u2245\u00a0\u2206CDA \u00a0 \u00a0 \u00a0[AAS-criterial]<br \/>\n<strong>(i)<\/strong> \u2206ABC \u2245 \u2206CDA \u00a0 \u00a0 (proved)<br \/>\n\u2234 AB = CD and BC = AD \u00a0 \u00a0 (c.p.c.t.)<br \/>\n(ii) \u2206ABC \u2245 \u2206CDA \u00a0 \u00a0 \u00a0(proved)<br \/>\n\u2234 \u2220B = \u2220D (c.p.c.t.)<br \/>\nAlso, \u22201 = \u22202 and \u22203 = \u22204<br \/>\n\u22201 + \u22204 = \u22202 + \u22203 \u21d2 \u2220A = \u2220C<br \/>\nHence, \u2220B = \u2220D and \u2220A = \u2220C<br \/>\n<strong>(iii)<\/strong> Since \u2206ABC \u2245 \u2206CDA and congruent triangles are equal in area,<br \/>\nSo we have \u2206ABC = \u2206CDA<br \/>\nSimilarly, \u2206ABD = \u2206CDB<\/p>\n<p><strong>Example 20: \u00a0 \u00a0<\/strong>If the diagonals of a parallelogram are perpendicular to each other, prove that it is a rhombus.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Since the diagonals of a ||gm bisect each other,<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8390\/29042741744_ebc4f448d9_o.png\" alt=\"Types-Of-Quadrilaterals-Example-20\" width=\"237\" height=\"167\" \/><br \/>\nwe have, OA = OC and OB = OD.<br \/>\nNow, in \u2206AOD and COD, we have<br \/>\nOA = OC, \u2220AOD = \u2220COD =90\u00b0<br \/>\nand OD is common<br \/>\n\u2234 \u2206AOD \u2245 \u2206COD<br \/>\n\u2234 AD = CD (c.p.c.t.)<br \/>\nNow, AB = CD and AD = BC<br \/>\n(opp. sides of a ||gm)<br \/>\nand AD = CD (proved)<br \/>\n\u2234 AB = CD = AD = BC<br \/>\nHence, ABCD is a rhombus.<\/p>\n<p><strong>Example 21: \u00a0 \u00a0<\/strong>PQRS is a square. Determine \u2220SRP.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0PQRS is a square.<br \/>\n\u2234 PS = SR and \u2220PSR = 90\u00b0<br \/>\nNow, in \u2206 PSR, we have<br \/>\nPS = SR<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8255\/29558126952_8dcbab93e7_o.png\" alt=\"Types-Of-Quadrilaterals-Example-21\" width=\"181\" height=\"179\" \/><br \/>\n\u21d2 \u22201 = \u22202<br \/>\n\u2235 Angles opposite to equal sides in a \u2206 are equal<br \/>\nBut, \u22201 + \u22202 + \u2220PSR = 180\u00ba<br \/>\n\u2234 2\u22201 + 90\u00b0 = 180\u00ba \u00a0 \u00a0 \u00a0[\u2235 \u2220PSR = 90\u00b0]<br \/>\n\u21d2 2\u22201 = 90\u00b0<br \/>\n\u21d2 \u22201 = 45\u00b0<\/p>\n<p><strong>Example 22: \u00a0 \u00a0<\/strong>In the adjoining figure, ABCD is a rhombus. If \u2220A = 70\u00ba, find \u2220CDB<br \/>\nSol.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8001\/29558126942_33c2a9dfba_o.png\" alt=\"Types-Of-Quadrilaterals-Example-22\" width=\"209\" height=\"158\" \/><br \/>\nWe have \u2220C = \u2220A = 70\u00ba<br \/>\n(opposite \u2220 of a ||gm)<br \/>\nLet \u2220CDB = x\u00ba<br \/>\nIn \u2206CDB, we have<br \/>\nCD = CB \u21d2\u2220CBD = \u2220CDB = x\u00ba<br \/>\n\u2234 \u2220CDB + \u2220CBD + \u2220DCB = 180\u00ba<br \/>\n(angles of a triangle)<br \/>\n\u21d2 x\u00ba + x\u00ba + 70\u00ba = 180\u00ba<br \/>\n\u21d2 2x = 110, i.e., x = 55<br \/>\nHence, \u2220CDB = 55\u00ba<\/p>\n<p><strong>Example 23: \u00a0 \u00a0<\/strong>ABCD is a rhombus with \u2220ABC = 56\u00b0. Determine \u2220ACD.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0ABCD is a parallelogram<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8240\/29042741344_90a7266df8_o.png\" alt=\"Types-Of-Quadrilaterals-Example-23\" width=\"227\" height=\"164\" \/><br \/>\n\u21d2 \u2220ABC = \u2220ADC<br \/>\n\u21d2 \u2220ADC = 56\u00b0 \u00a0 \u00a0 [\u2235 \u2220ABC = 56\u00b0 (Given)]<br \/>\n\u21d2 \u2220ODC = 28\u00b0 \u00a0 \u00a0[\u2235 \u2220ODC = 1\/2 \u2220ADC]<br \/>\nNow, \u2206OCD we have,<br \/>\n\u2220OCD + \u2220ODC + \u2220COD = 180\u00b0<br \/>\n\u21d2 \u2220ODC + 28\u00b0 + 90\u00b0 = 180\u00b0<br \/>\n\u21d2 \u2220OCD = 62\u00b0 \u21d2\u2220ACD = 62\u00b0.<\/p>\n<p><strong>Example 24: \u00a0 \u00a0<\/strong>Prove that the line segment joining the<br \/>\nmid-points of the diagonals of a trapezium is parallel to each of the parallel sides and is equal to half the difference of these sides.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>Given:<\/strong> A trapezium ABCD in which<br \/>\nAB || DC and P and Q are the mid-points of its diagonals AC and BD respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8238\/29378302570_fdf2416fb3_o.png\" alt=\"Types-Of-Quadrilaterals-Example-24\" width=\"210\" height=\"188\" \/><br \/>\n<strong>To Prove:<\/strong> (i) PQ || AB or DC<br \/>\n(ii) PQ = 1\/2 (AB \u2013 DC)<br \/>\n<strong>Construction:<\/strong> Join DP and produce DP to meet AB in R.<br \/>\n<strong>Proof:<\/strong> Since AB || DC and transversal AC cuts them at A and C respectively.<br \/>\n\u22201 = \u22202 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;. (i)<br \/>\n[\u2234 Alternate angles are equal]<br \/>\nNow, in \u2206s APR and DPC, we have<br \/>\n\u22201 = \u22202 \u00a0 \u00a0 \u00a0 \u00a0[From (i)]<br \/>\nAP = CP \u00a0 \u00a0 \u00a0 [\u2235 P is the mid-point of AC]<br \/>\nand, \u22203 = \u22204 \u00a0 \u00a0 \u00a0 \u00a0[Vertically opposite angles]<br \/>\nSo, by ASA criterion of congruence<br \/>\n\u2206APR \u2245 \u2206DPC<br \/>\n\u21d2 AR = DC and PR = DP &#8230;.(ii)<br \/>\n[\u2235 Corresponding parts of congruent triangles are equal]<br \/>\nIn \u2206DRB, P and Q are the mid-points of sides DR and DB respectively.<br \/>\n\u2234 PQ || RB<br \/>\n\u21d2 PQ || AB [\u2235 RB is a part of AB]<br \/>\n\u21d2 PQ || AB and DC \u00a0 \u00a0 \u00a0[\u2235 AB || DC (Given)]<br \/>\nThis proves (i).<br \/>\nAgain, P and Q are the mid-points of sides DR and DB respectively in \u2206DRB.<br \/>\n\u2234 PQ = 1\/2 RB \u21d2 PQ = 1\/2 (AB \u2013 AR)<br \/>\n\u21d2 PQ = 1\/2 (AB \u2013 DC) \u00a0 \u00a0 \u00a0 [From (ii), AR = DC]<br \/>\nThis proves (ii).<\/p>\n<p><strong>Example 25: \u00a0 \u00a0<\/strong>In the adjoining figure, ABCD is parallelogram and X, Y are the points on diagonal BD such that DX = BY. Prove that CXAY is a parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8213\/29042741084_cdfa32fbeb_o.png\" alt=\"Types-Of-Quadrilaterals-Example-25\" width=\"280\" height=\"163\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Join AC, meeting BD at O.<br \/>\nSince the diagonals of a parallelogram bisect each other, we have OA = OC and OD = OB.<br \/>\nNow, OD = OB and DX = BY<br \/>\n\u21d2 OD \u2013 DX = OB \u2013 BY \u21d2 OX = OY<br \/>\nNow, OA = OC and OX = OY<br \/>\n\u2234 CXAY is a quadrilateral whose diagonals bisect each other.<br \/>\n\u2234 CXAY is a ||gm<\/p>\n<p><strong>Example 26: \u00a0 \u00a0<\/strong>Prove that the four triangles formed by joining in pairs, the mid-points of three sides of a triangle, are concurrent to each other.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>Given:<\/strong> A triangle ABC and D,E,F are the mid-points of sides BC, CA and AB respectively.<br \/>\n<strong>To Prove:<\/strong><br \/>\n\u2206 AFE \u2245 \u2206FBD \u2245 \u2206EDC \u2245 \u2206DEF.<br \/>\n<strong>Proof:<\/strong> Since the segment joining the mid-points of the sides of a triangle is half of the third side. Therefore,<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8881\/29633520686_011a2ccfcd_o.png\" alt=\"Types-Of-Quadrilaterals-Example-26\" width=\"173\" height=\"181\" \/><br \/>\nDE = 1\/2 \u21d2 DE = AF = BF \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;. (i)<br \/>\nEF = 1\/2 \u21d2 EF = BD = CD \u00a0 \u00a0 \u00a0 \u00a0 &#8230;. (ii)<br \/>\nDF = 1\/2 \u21d2 DF = AE = EC \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(iii)<br \/>\nNow, in \u2206s DEF and AFE, we have<br \/>\nDE = AF \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[From (i)]<br \/>\nDF = AE \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [From (ii)]<br \/>\nand, EF = FE \u00a0 \u00a0[Common]<br \/>\nSo, by SSS criterion of congruence,<br \/>\n\u2206 DEF \u2245 \u2206AFE<br \/>\nSimilarly, \u2206DEF \u2245 \u2206FBD and \u2245 DEF \u2245 \u2206EDC<br \/>\nHence, \u2206 AFE \u2245 \u2206FBD \u2245 \u2206EDC \u2245 \u2206DEF.<\/p>\n<p><strong>Example 27: \u00a0 \u00a0<\/strong>In fig, AD is the median and DE || AB. Prove that BE is the median.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0In order to prove that BE is the median, it is sufficient to show that E is the mid-point of AC.<br \/>\nNow, AD is the median in \u2206ABC<br \/>\n\u21d2 D is the mid-point of BC.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8396\/29378302310_a6b7ee2607_o.png\" alt=\"Types-Of-Quadrilaterals-Example-27\" width=\"213\" height=\"162\" \/><br \/>\nSince DE is a line drawn through the mid-point of side BC of \u2206ABC and is parallel to AB (given). Therefore, E is the mid-point of AC. Hence, BE is the median of \u2206ABC.<\/p>\n<p><strong>Example 28: \u00a0 \u00a0<\/strong>Let ABC be an isosceles triangle with\u00a0AB = AC and let D,E,F be the mid-points of BC, CA and AB respectively. Show that<br \/>\nAD \u22a5 FE and AD is bisected by FE.<br \/>\n<strong>Solution: \u00a0 \u00a0Given:<\/strong> An isosceles triangle ABC with D, E and F as the mid-points of sides BC, CA and AB respectively such that AB = AC. AD intersects FE at O.<br \/>\n<strong>To Prove:<\/strong> AD \u22a5\u00a0FE and AD is bisected by FE.<br \/>\n<strong>Constructon:<\/strong> Join DE and DF.<br \/>\n<strong>Proof:<\/strong> Since the segment joining the\u00a0mid-points of two sides of a triangle is parallel to third side and is half of it. Therefore,<br \/>\nDE || AB and DE = 1\/2 AB<br \/>\nAlso, DF || AC and DF = 1\/2 AC<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8399\/29042740794_4745de82cb_o.png\" alt=\"Types-Of-Quadrilaterals-Example-28\" width=\"201\" height=\"200\" \/><br \/>\nBut, 1\/2 AB = 1\/2 AC \u00a0 \u00a0[Given]<br \/>\n\u21d2 AB = AC<br \/>\n\u21d2 DE = DF \u00a0 \u00a0 \u00a0 &#8230;. (i)<br \/>\nNow, DE = 1\/2 AB \u21d2 DE = AF \u00a0 \u00a0 \u00a0 &#8230;. (ii)<br \/>\nand, DF = 1\/2 AC \u21d2 DF = AE \u00a0 \u00a0 \u00a0 \u00a0 &#8230;(iii)<br \/>\nFrom (i), (ii) and (iii) we have<br \/>\nDE = AE = AF = DF<br \/>\n\u21d2 DEAF is a rhombus.<br \/>\n\u21d2 Diagonals AD and FE bisect each other at right angle.<br \/>\nAD \u22a5 FE and AD is bisected by FE.<\/p>\n<p><strong>Example 29: \u00a0 \u00a0<\/strong>ABCD is a parallelogram. P is a point on AD such that AP = 1\/3 AD and Q is a point on BC such that CQ = 1\/3 BP. Prove that AQCP is a parallelogram.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8002\/29633520606_2738f043cf_o.png\" alt=\"Types-Of-Quadrilaterals-Example-29\" width=\"219\" height=\"169\" \/><br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0ABCD is a parallelogram.<br \/>\n\u21d2 AD = BC and AD || BC<br \/>\n\u21d2 1\/3 AD = 1\/3 BC and AD || BC<br \/>\n\u21d2 AP = CQ and AP || CQ<br \/>\nThus, APCQ is a quadrilateral such that one pair of opposite side AP and CQ are parallel and equal.<br \/>\nHence, APCQ is a parallelogram.<\/p>\n<p><strong>Example 30: \u00a0 \u00a0<\/strong>In fig. D,E and F are, respectively the mid- points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle.<br \/>\n<strong>Solution: \u00a0<\/strong> \u00a0Since the segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8351\/29378302050_e6940232ee_o.png\" alt=\"Types-Of-Quadrilaterals-Example-30\" width=\"210\" height=\"202\" \/><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2835 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-30-1.png\" alt=\"types-of-quadrilaterals-example-30-1\" width=\"428\" height=\"466\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-30-1.png 428w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-30-1-276x300.png 276w\" sizes=\"(max-width: 428px) 100vw, 428px\" \/><\/p>\n<p><strong>Example 31: \u00a0 \u00a0<\/strong>P,Q and R are, respectively, the mid-points of sides BC, CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. Prove that XY = 1\/4 BC<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>Given:<\/strong> A \u2206ABC with P,Q and R as the\u00a0mid-points of BC, CA and AB respectively. PR and BQ meet at X and CR and PQ meet at Y.<br \/>\n<strong>Construction:<\/strong> Join \u201cX and Y.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8151\/29633520466_60e86a0696_o.png\" alt=\"Types-Of-Quadrilaterals-Example-31\" width=\"231\" height=\"198\" \/><br \/>\n<strong>Proof:<\/strong> Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it. Therefore, Q and R are mid-points of AC and AB respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2836 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-31-1.png\" alt=\"types-of-quadrilaterals-example-31-1\" width=\"437\" height=\"339\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-31-1.png 437w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-31-1-300x233.png 300w\" sizes=\"(max-width: 437px) 100vw, 437px\" \/><br \/>\nSimilarly, Y is the mid-point of PQ.<br \/>\nNow, consider \u2206PQR. XY is the line segment joining the mid-points of sides PR and PQ.<br \/>\n\u2234 XY = 1\/2 RQ \u00a0 \u00a0 \u00a0&#8230;. (i)<br \/>\nBut RQ = 1\/2 BC \u00a0 \u00a0[From (i)]<br \/>\nHence, XY = 1\/4 BC<\/p>\n<p><strong>Example 32: \u00a0 \u00a0<\/strong>Show that the quadrilateral, formed by joining the mid-points of the sides of a square, is also a square.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>Given:<\/strong> A square ABCD in which P, Q, R, S are the mid-points of sides AB, BC, CD, DA respectively. PQ, QR, RS and SP are joined.<br \/>\n<strong>To Prove:<\/strong> PQRS is a square.<br \/>\n<strong>Construction:<\/strong> Join AC and BD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8239\/29042740404_3dcc065d8f_o.png\" alt=\"Types-Of-Quadrilaterals-Example-32\" width=\"213\" height=\"235\" \/><br \/>\nProof : In \u2206ABC, P and Q are the mid-points of sides AB and BC respectively.<br \/>\n\u2234 PQ || AC and PQ = 1\/2 AC \u00a0 \u00a0 &#8230;. (i)<br \/>\nIn \u2206ADC, R and S are the mid-points of CD and AD respectively.<br \/>\n\u2234 RS || AC and RS = 1\/2 AC \u00a0 \u00a0 \u00a0 &#8230;.(ii)<br \/>\nFrom (i) and (ii), we have<br \/>\nPQ || RS and PQ = RS \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(iii)<br \/>\nThus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.<br \/>\nHence, PQRS is a parallelogram.<br \/>\nNow, in \u2206s PBQ and RCQ, we have<br \/>\nPB = RC<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2837 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-32-1.png\" alt=\"types-of-quadrilaterals-example-32-1\" width=\"362\" height=\"465\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-32-1.png 362w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Types-Of-Quadrilaterals-Example-32-1-234x300.png 234w\" sizes=\"(max-width: 362px) 100vw, 362px\" \/><br \/>\nNow, PQ || AC \u00a0 \u00a0 \u00a0[From (i)]<br \/>\n\u21d2 PM || NO \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(vi)<br \/>\nSince P and S are the mid-points of AB and AD respectively.<br \/>\nPS || BD<br \/>\n\u21d2 PM || MO \u00a0 \u00a0 \u00a0&#8230;.(vii)<br \/>\nThus, in quadrilateral PMON, we have<br \/>\nPM || NO \u00a0 \u00a0 \u00a0 [From (vi)]<br \/>\nPN || MO \u00a0 \u00a0 \u00a0 [From (vii)]<br \/>\nSo, PMON is a parallelogram.<br \/>\n\u21d2 \u2220MPN = \u2220MON<br \/>\n\u21d2 \u2220MPN = \u2220BOA \u00a0 \u00a0 \u00a0[\u2235 \u2220MON = \u2220BOA]<br \/>\n\u21d2 \u2220MPN = 90\u00b0<br \/>\n\u21d2 \u2220QPS = 90\u00b0<br \/>\nThus, PQRS is a quadrilateral such that<br \/>\nPQ = QR = RS = SP and \u2220QPS = 90\u00b0.<br \/>\nHence, PQRS is a square.<\/p>\n<p><strong>Example 33: \u00a0 \u00a0<\/strong>\u2206ABC is a triangle right angled at B ; and P is the mid-point of AC. Prove that PB = PA = 1\/2 AC.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0Given : \u2206ABC right angled at B, P is the mid-point of AC.<br \/>\n<strong>To Prove:<\/strong> PB = PA = 1\/2 AC.<br \/>\n<strong>Construction:<\/strong> Through P draw PQ || BC meeting AB at Q.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8531\/29042740174_aaecfcaaec_o.png\" alt=\"Types-Of-Quadrilaterals-Example-33\" width=\"211\" height=\"189\" \/><br \/>\n<strong>Proof:<\/strong> Since PQ || BC. Therefore,<br \/>\n\u2220AQP = \u2220ABC \u00a0 \u00a0 \u00a0[Corresponding angles]<br \/>\n\u21d2 \u2220AQP = 90\u00b0 \u00a0 \u00a0 \u00a0 [\u2235 \u2220ABC = 90\u00b0]<br \/>\nBut, \u2220AQP + \u2220BQP = 180\u00b0<br \/>\n[\u2235 \u2220AQP &amp; \u2220BQP are angles of a linear pair]<br \/>\n\u2234 90\u00b0 + \u2220BQP = 180\u00b0<br \/>\n\u21d2 \u2220BQP = 90\u00b0<br \/>\nThus, \u2220AQP = \u2220BQP = 90\u00b0<br \/>\nNow, in \u2206ABC, P is the mid-point of AC and<br \/>\nPQ || BC. Therefore, Q is the mid-point of AB i.e, AQ = BQ.<br \/>\nConsider now \u2206s APQ and BPQ.<br \/>\nwe have, AQ = BC [Proved above]<br \/>\n\u2220AQP = \u2220BQP [From (i)]<br \/>\nand, PQ = PQ<br \/>\nSo, by SAS cirterion of congruence<br \/>\n\u2206APQ \u2245 \u2220BPQ<br \/>\n\u21d2 PA = PB<br \/>\nAlso,<br \/>\nPS = 1\/2 AC, since P is the mid-point of AC<br \/>\nHence, PA = PB = 1\/2 AC.<\/p>\n<p><strong>Example 34: \u00a0 \u00a0<\/strong>Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0<strong>Given:<\/strong> A rectangle ABCD in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.<br \/>\n<strong>To Prove:<\/strong> PQRS is rhombus.<br \/>\n<strong>Construction:<\/strong> Join AC.<br \/>\n<strong>Proof:<\/strong> In \u2206ABC, P and Q are the mid-points of sides AB and BC respectively.<br \/>\n\u2234 PQ || AC and PQ = AC \u00a0 \u00a0 \u00a0&#8230;. (i)<br \/>\nIn \u2206 ADC, R and S are the mid-points of CD and AD respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/c1.staticflickr.com\/9\/8030\/29633520376_3bec0d30a9_o.png\" alt=\"Types-Of-Quadrilaterals-Example-34\" width=\"284\" height=\"192\" \/><br \/>\n\u2234 SR || AC and SR = 1\/2 AC \u00a0 \u00a0&#8230;. (ii)<br \/>\nFrom (i) and (ii), we get<br \/>\nPQ || SR and PQ = SR \u00a0 \u00a0 \u00a0 &#8230;.(iii)<br \/>\n\u21d2 PQRS is a parallelogram.<br \/>\nNow, ABCD is a rectangle.<br \/>\n\u21d2 AD = BC \u21d2 1\/2 AD = 1\/2 BC<br \/>\n\u21d2 AS = BQ \u00a0 \u00a0 \u00a0 &#8230;.(iv)<br \/>\nIn \u2206s APS and BPQ , we have<br \/>\nAP = BP \u00a0 \u00a0 \u00a0 [\u2234P is the mid-point of AB]<br \/>\n\u2220PAS = \u2220PBQ \u00a0 \u00a0 \u00a0[Each equal to 90\u00b0]<br \/>\nand, AS = BQ \u00a0 \u00a0 \u00a0 \u00a0[From (iv)]<br \/>\nSo, by SAS criterion of congruence<br \/>\n\u2206APS \u2245 \u2206BPQ<br \/>\nPS = PQ \u00a0 \u00a0 \u00a0 &#8230;.(v)<br \/>\n[\u2235 Corresponding parts of congruent\u00a0triangles are equal]<br \/>\nFrom (iii) and (v), we obtain that PQRS is a parallelogram such that PS = PQ i.e., two adjacent sides are equal.<br \/>\nHence, PQRS is a rhombus.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How Many Types Of Quadrilaterals Are There A quadrilateral is a figure bounded by four line segments such that no three of them are parallel. Two sides of quadrilateral are consecutive or adjacent sides, if they have a common point (vertex). Two sides of a quadrilateral are opposite sides, if they have no common end-point [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[1212,1218,1210,1215,1216,1214,1217,1213,1211],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>How Many Types Of Quadrilaterals Are There - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/types-of-quadrilaterals\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How Many Types Of Quadrilaterals Are There\" \/>\n<meta property=\"og:description\" content=\"How Many Types Of Quadrilaterals Are There A quadrilateral is a figure bounded by four line segments such that no three of them are parallel. 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Two sides of quadrilateral are consecutive or adjacent sides, if they have a common point (vertex). 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