**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 (vertex).
- The consecutive angles of a quadrilateral are two angles which include a side in their intersection.
- In other words, two angles are consecutive, if they have a common arm.
- Two angles of a quadrilateral are said to be opposite angles if they do not have a common arm.
- The sum of the four angles of a quadrilateral is 360º.
- A quadrilateral having exactly one pair of parallel sides, is called a trapezium.
- A trapezium is said to be an isoscels trapezium, if its non-parallel sides are equal.
- A quadrilateral is a parallelogram if its both pairs of opposite sides are parallel.
- A parallelogram having all sides equal is called a rhombus.
- A parallelogram whose each angle is a right angle, is called a rectangle.
- A square is a rectangle with a pair of adjacent sides equal.
- A quadrilateral is a kite if it has two pairs of equal adjacent sides and unequal opposite sides.
- A diagonal of a parallelogram divides it into two congruent triangles.
- In a parallelogram, opposite sides are equal.
- The opposite angles of a parallelogram are equal.
- The diagonals of a parallelogram bisect each other.
- In a parallelogram, the bisectors of any two consecutive angles intersect at right angle.
- If diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle.
- The angle bisectors of a parallegram form a rectangle.
- A quadrilateral is a parallelogam if its opposite sides are equal.
- A quadrilateral is a parallelogram if its opposite angles are equal.
- If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
- A quadrilateral is a parallelogram, if its one pair of opposite sides are equal and parallel.
- Each of the four angles of a rectangel is a right angle.
- Each of the four sides of a rhombus is of the same length.
- Each of the angles of a square is a right angle and each of the four sides is of the same length.
- The diagonals of a rectangle are of equal length.
- If the two diagonals of parallelogram are equal, it is a rectangle.
- The diagonals of a rhombus are perpendicular to each other.
- If the diagonals of a parallelogram are perpendicular, then it is a rhombus.
- The diagonals of a square are equal and perpendicular to each other.
- If the diagonals of a parallelogram are equal and intersect at right angles then the parallelogram is a square.
- A diagonal of a parallelogram divides it into two triangles of equal area.
- 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.
- Parallelograms on the same base and between the same parallels are equal in area.
- A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
- The area of a parallelogram is the product of its base and the corresponding altitude.
- Parallelograms on equal bases and between the same parallels are equal in area.

**Read More: **

- Different Kinds of Quadrilateral
- Properties of Cyclic Quadrilaterals
- RS Aggarwal Class 9 Solutions Quadrilaterals and Parallelograms

**Types Of Quadrilaterals Example Problems With Solutions**

**Example 1: **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.

**Solution:** We have ∠A : ∠B : ∠C : ∠D = 2 : 4 : 5 : 7.

So, let ∠A = 2xº, ∠B = 4xº, ∠C = 5xº, ∠D = 7xº.

∴ ∠A + ∠B + ∠C + ∠D = 360º

⇒ x + 4x + 5x + 7x = 360º

⇒ 18x = 360º

⇒ x = 20º

Thus, the angles are:

∠A = 40º, ∠B = (4 × 20)º = 80º,

∠C = (5 × 20)º = 100º

and, ∠D = (7x)º = (7 × 20)º = 140º

**Example 2: **The sides BA and DC of a quadrilateral ABCD are produced as shown in fig.

Prove that a + b = x + y.

**Solution:** Join BD. In ∆ABD, we have

∠ABD + ∠ADB = bº ….(i)

In ∆CBD, we have

∠CBD + ∠CDB = aº ….(ii)

Adding (i) and (ii), we get

(∠ABD + ∠CBD) + (∠ADB + ∠CDB) = aº + bº

⇒ xº + yº = aº + bº

Hence, x + y = a + b

**Example 3: **In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B respectively. Prove that ∠AOB = 1/2 (∠C + ∠D).

**Solution:** In ∆AOB, we have

∠AOB + ∠1 + ∠2 = 180º

**Example 4: **In figure bisectors of ∠B and ∠D of quadrilateral ABCD meet CD and AB produced at P and Q respectively. Prove that

∠P + ∠Q = 1/2 (∠ABC + ∠ADC)

**Solution:**

**Example 5: **In a parallelogram ABCD, prove that sum of any two consecutive angles is 180º.

**Solution:** Since ABCD is a parallelogram. Therefore,

AD || BC.

Now, AD || BC and transversal AB intersects them at A and B respectively.

∴ ∠A + ∠B = 180º

[∵ Sum of the interior angles on the same side of the transversal is 180º]

Similarly, we can prove that

∠B + ∠C = 180º, ∠C + ∠D = 180º and

∠D + ∠A = 180º.

**Example 6: **In a parallelogram ABCD, ∠D = 115º, determine the measure of ∠A and ∠B.

**Solution:** Since the sum of any two consecutive angles of a parallelogram is 180º. Therefore,

∠A + ∠D = 180º and ∠A + ∠B = 180º

Now, ∠A + ∠D = 180º

⇒ ∠A + 115º = 180º [∵ ∠D = 115º (given)]

⇒ ∠A = 65º and ∠A + ∠B = 180º

⇒ 65º + ∠B = 180º ⇒ ∠B = 115º

Thus, ∠A = 65º and ∠B = 115º

**Example 7: **In figure, AB = AC, ∠EAD = ∠CAD and

CD || AB. Show that ABCD is a parallelogram.

**Solution:** In ∆ABC, AB = AC [Given]

⇒ ∠ABC = ∠ACB ….(1)

(Angles opposite the equal sides are equal)

∠EAD = ∠CAD[Given] ….(2)

Now, ∠EAC = ∠ABC + ∠ACB

[An exterior angle is equal to sum of two interior opposite angles of a triangles]

⇒ ∠EAD + ∠CAD = ∠ABC + ∠ACB

⇒ ∠CAD + ∠CAD = ∠ACB + ∠ACB

By (1) and (2)

⇒ 2∠CAD = 2∠ACB

⇒ ∠CAD = ∠ACB

⇒ BC | | AD

Also, CD | | AB [Given]

Thus, we have both pairs of opposite sides of quadrilateral ABCD parallel. Therefore, ABCD is a parallelogram.

**Example 8: **ABCD is a parallelogram and line segments AX,CY are angle bisector of ∠A and ∠C respectively then show AX || CY.

**Solution:** Since opposite angles are equal in a parallelogram. Therefore, in parallelogram ABCD, we have ∠A = ∠C

⇒ 1/2 ∠A = 1/2 ∠C

⇒ ∠1 = ∠2 ….(i)

[∵ AX and CY are bisectors of ∠A and ∠C respectively]

Now, AB || DC and the transversal CY intersects them.

∴ ∠2 = ∠3 …(ii)

[∵ Alternate interior angles are equal]

From (i) and (ii), we get

∠1 = ∠3

Thus, transversal AB intersects AX and YC at A and Y such that ∠1 = ∠3 i.e. corresponding angles are equal.

∴ AX || CY

**Example 9: **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.

**Solution:** Given a quad. ABCD in which AB = BC

= CD = DA and O is a point within it such that OB = OD.

To prove ∠AOB + ∠COB = 180º

Proof In ∆OAB and OAD, we have

AB = AD (given)

OA = OA (common)

and OB = OD (given)

∴ ∆OAB ≅ ∆OAD

∴ ∠AOB = ∠AOD ….(i) (c.p.c.t.)

Similarly, ∆OBC ≅ ∆ODC

∴∠COB = ∠COD ….(ii)

Now, ∠AOB + ∠COB + ∠COD + ∠AOD

= 360º [∠ at a point]

⇒ 2(∠AOB + ∠COB) = 360º

⇒ ∠AOB + ∠COB = 180º

**Example 10: **In figure AN and CP are perpendiculars to the diagonal BD of a parallelogram ABCD. Prove that:

(i) ∆ADN ≅ ∆CBP (ii) AN = CP

**Solution:** Since ABCD is a parallelogram.

∴ AD || BC

Now, AD || BC and transversal BD intersects them at B and D.

∴ ∠1 = ∠2

[∵ Alternate interior angles are equal]

Now, in ∆s ADN and CBP, we have

∠1 = ∠2

∠AND = ∠CPD and, AD = BC

[∵ Opposite sides of a ||gm are equal]

So, by AAS criterion of congruence

∆ADN ≅ ∆CBP

AN = CP

[∵ Corresponding parts of congruent triangles are equal]

**Example 11: **In figure, ABCD is a trapezium such that AB || CD and AD = BC.

BE || AD and BE meets BC at E.

Show that (i) ABED is a parallelogram.

(ii) ∠A + ∠C = ∠B + ∠D = 180º.

**Solution:** Here, AB || CD (Given)

⇒ AB || DE ….(1)

Also, BE || AD (Given) ….(2)

From (1) and (2),

ABED is a parallelogram

⇒ AD = BE ….(3)

Also, AD = BC (Given) ….(4)

From (3) and (4),

BE = BC

⇒ ∠BEC = ∠BCE ….(5)

Also, ∠BAD = ∠BED

(opposite angles of parallelogram ABED)

i.e., ∠BED = ∠BAD ….(6)

Now, ∠BED + ∠BEC = 180º (Linear pair of angles)

⇒ ∠BAD + ∠BCE = 180º

By (5) and (6)

⇒ ∠A + ∠C = 180º

Similarly, ∠B + ∠D = 180º

**Example 12: **In figure ABCD is a parallelogram and ∠DAB = 60º. 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.

**Solution:** We have, ∠DAB = 60º

∠A + ∠B = 180º

∴ 60º + ∠B = 180º ⇒ ∠B = 120º

Now, AB || DC and transversal AP intersects them.

∴ ∠PAB = ∠APD

⇒ ∠APD = 30º [∵ ∠PAB = 30º]

Thus, in ∆APD, we have

∠PAD = ∠APD [Each equal to 30º]

⇒ AD = PD …. (i)

[∵ Angles opposite to equal sides are equal] Since BP is the bisector of ∠B. Therefore,

∠ABP = ∠PBC = 60º

Now, AB || DC and transversal BP intersects them.

∴ ∠CPB = ∠ABP

⇒ ∠CPB = 60º [∵ ∠ABP = 60º]

Thus, in ∆CBP, we have

∠CBP = ∠CPB [Each equal to 60º]

⇒ CP = BC

∵ [Sides opp, to equal angles are equal]

⇒ CP = AD …. (ii)

[∵ ABCD is a parallelogram ∴ AD = BC]

From (i) and (ii), we get

PD = CP

⇒ P is the mid point of CD.

**Example 13: **Prove that the line segments joining the mid-point of the sides of a quadrilateral forms a parallelogram.

**Solution:** Points 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.

Join the diagonal AC of the quadrilateral ABCD.

Now, in ∆ABC, we have E and F mid-points of the sides BA and BC.

⇒ EF || AC

and EF = 1/2 AC …. (1)

Similarly, from ∆ADC, we have

GH || AC

and GH = 1/2 AC ….(2)

Then from (1) and (2), we have

EF || GH

and EF = GH

This proves that EFGH is a parallelogram.

**Example 14: **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.

**Solution:** Since X and Y are the mid-points of AB and DC respectively. Therefore,

AX = 1/2 AB and CY = 1/2 DC … (i)

But, AB = DC [∵ ABCD is a parallelogram]

⇒ 1/2 AB = 1/2 DC

⇒ AX = CY …. (ii)

Also, AB || DC

⇒ AX || YC …. (iii)

Thus, in quadrilateral AXCY, we have

AX || YC and AX = YC

[From (ii) and (iii)]

Hence, quadrilateral AXCY is a parallelogram.

**Example 15: **Prove that the line segments joining the mid-points of the sides of a rectangle forms a rhombus.

**Solution:** P, Q, R and S are the mid-points of the sides AB, BC, CD and DA of the rectangle ABCD.

**Example 16: **In figure ABCD is a parallelogram and X and Y are points on the diagonal BD such that

DX = BY. Prove that

(i) AXCY is a parallelogram

(ii) AX = CY, AY = CX

(iii) ∆AYB ≅ ∆CXD

**Solution:** Given : ABCD is a parallelogram. X and Y are points on the diagonal BD such that

DX = BY

**To Prove:**

(i) AXCY is a parallelogram

(ii) AX = CY, AY = CX

(iii) ∆AYB ≅ ∆CXD

C Construction : join AC to meet BD at O.

**Proof:**

**(i)** We know that the diagonals of a parallelogram bisect each other. Therefore, AC and BD bisect each other at O.

∴ OB = OD

But, BY = DX

∴ OB – BY = OD – DX

⇒ OY = OX

Thus, 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.

Hence, AXCY is a parallelogram.

**(ii)** Since AXCY is a parallelogram

∴ AX = CY and AY = CX

**(iii)** In triangles AYB and CXD, we have

AY = CX [From (ii)]

AB = CD

[∵ ABCD is a parallelogram]

BY = DX [Given]

So, by SSS-criterion of congruence, we have

∆AYB ≅ ∆CXD

**Example 17: **In fig. ABC is an isosceles triangle in which AB = AC. CP || AB and AP is the bisector

of exterior ∠CAD of ∆ABC. Prove that

∠PAC = ∠BCA and ABCP is a parallelogram.

**Solution:** Given : An isosceles ∆ABC having

AB = AC.AP is the bisector of ext ∠CAD and CP || AB.

**To Prove:** ∠PAC = ∠BCA and ABCP

**Proof:** In ∆ABC, we have

AB = AC [Given]

⇒ ∠1 = ∠2 …. (i)

∵ Angles opposite to equal sides in a ∆ are equal

Now, in ∆ ABC, we have

ext ∠CAD = ∠1 + ∠2

⇒ ext ∠CAD = 2∠2 [∵ ∠1 = ∠2 (from (i))]

⇒ 2∠3 = 2∠2

[∵ AP is the bisector of ext.∠CAD ∴∠CAD = 2∠3]

⇒ ∠3 = ∠2

Thus, AC intersects lines AP and BC at A and C respectively such that ∠3 = ∠2

i.e., alternate interior angles are equal.

Therefore, AP || BC.

But, CP || AB [Gvien]

Thus, ABCP is a quadrilateral such that

AP || BC and CP || AB.

Hence, ABCP is a parallelogram.

**Example 18: **In the given figure, ABCD is a square and ∠PQR = 90º. If PB = QC = DR, prove that

(i) QB = RC, (ii) PQ = QR, (iii) ∠QPR = 45º.

**Solution:** BC = DC, CQ = DR ⇒ BC – CQ = ∆CDR

⇒ QB = RC

From ∆CQR, ∠RQB = ∠QCR + ∠QRC

⇒ ∠RQP + ∠PQB = 90º + ∠QRC

⇒ 90º + ∠PQB = 90º + ∠QRC

Now, ∆RCQ ≅ ∆QBP and therefore,

QR = PQ

PQ = QR ⇒ ∠QPR = ∠PRQ

Bur, ∠QPR + ∠PRQ = 90º.

So, ∠QPR = 45º

**Example 19: **Prove that in a parallelogram

(i) opposite sides are equal

(ii) opposite angles are equal

(iii) each diagonal bisects the parallelogram

**Solution:** **Given:** A ||gm ABCD in which AB || DC and AD || BC.

**To prove:** (i) AB = CD and BC = AD;

(ii) ∠B = ∠D and ∠A = ∠C,

(iii) ∆ABC = ∆CDA and ∆ABD = ∆CDB

Construction join A and C.

In ∆ABC and CDA, we have,

∠1 = ∠2

[Alt. int. ∠, as AB || DC and CA cuts them]

∠3 = ∠4

[Alt. int. ∠, as BC || AD and CA cuts them]

AC = CA (common)

∴ ∆ABC ≅ ∆CDA [AAS-criterial]

**(i)** ∆ABC ≅ ∆CDA (proved)

∴ AB = CD and BC = AD (c.p.c.t.)

(ii) ∆ABC ≅ ∆CDA (proved)

∴ ∠B = ∠D (c.p.c.t.)

Also, ∠1 = ∠2 and ∠3 = ∠4

∠1 + ∠4 = ∠2 + ∠3 ⇒ ∠A = ∠C

Hence, ∠B = ∠D and ∠A = ∠C

**(iii)** Since ∆ABC ≅ ∆CDA and congruent triangles are equal in area,

So we have ∆ABC = ∆CDA

Similarly, ∆ABD = ∆CDB

**Example 20: **If the diagonals of a parallelogram are perpendicular to each other, prove that it is a rhombus.

**Solution:** Since the diagonals of a ||gm bisect each other,

we have, OA = OC and OB = OD.

Now, in ∆AOD and COD, we have

OA = OC, ∠AOD = ∠COD =90°

and OD is common

∴ ∆AOD ≅ ∆COD

∴ AD = CD (c.p.c.t.)

Now, AB = CD and AD = BC

(opp. sides of a ||gm)

and AD = CD (proved)

∴ AB = CD = AD = BC

Hence, ABCD is a rhombus.

**Example 21: **PQRS is a square. Determine ∠SRP.

**Solution:** PQRS is a square.

∴ PS = SR and ∠PSR = 90°

Now, in ∆ PSR, we have

PS = SR

⇒ ∠1 = ∠2

∵ Angles opposite to equal sides in a ∆ are equal

But, ∠1 + ∠2 + ∠PSR = 180º

∴ 2∠1 + 90° = 180º [∵ ∠PSR = 90°]

⇒ 2∠1 = 90°

⇒ ∠1 = 45°

**Example 22: **In the adjoining figure, ABCD is a rhombus. If ∠A = 70º, find ∠CDB

Sol.

We have ∠C = ∠A = 70º

(opposite ∠ of a ||gm)

Let ∠CDB = xº

In ∆CDB, we have

CD = CB ⇒∠CBD = ∠CDB = xº

∴ ∠CDB + ∠CBD + ∠DCB = 180º

(angles of a triangle)

⇒ xº + xº + 70º = 180º

⇒ 2x = 110, i.e., x = 55

Hence, ∠CDB = 55º

**Example 23: **ABCD is a rhombus with ∠ABC = 56°. Determine ∠ACD.

**Solution:** ABCD is a parallelogram

⇒ ∠ABC = ∠ADC

⇒ ∠ADC = 56° [∵ ∠ABC = 56° (Given)]

⇒ ∠ODC = 28° [∵ ∠ODC = 1/2 ∠ADC]

Now, ∆OCD we have,

∠OCD + ∠ODC + ∠COD = 180°

⇒ ∠ODC + 28° + 90° = 180°

⇒ ∠OCD = 62° ⇒∠ACD = 62°.

**Example 24: **Prove that the line segment joining the

mid-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.

**Solution:** **Given:** A trapezium ABCD in which

AB || DC and P and Q are the mid-points of its diagonals AC and BD respectively.

**To Prove:** (i) PQ || AB or DC

(ii) PQ = 1/2 (AB – DC)

**Construction:** Join DP and produce DP to meet AB in R.

**Proof:** Since AB || DC and transversal AC cuts them at A and C respectively.

∠1 = ∠2 …. (i)

[∴ Alternate angles are equal]

Now, in ∆s APR and DPC, we have

∠1 = ∠2 [From (i)]

AP = CP [∵ P is the mid-point of AC]

and, ∠3 = ∠4 [Vertically opposite angles]

So, by ASA criterion of congruence

∆APR ≅ ∆DPC

⇒ AR = DC and PR = DP ….(ii)

[∵ Corresponding parts of congruent triangles are equal]

In ∆DRB, P and Q are the mid-points of sides DR and DB respectively.

∴ PQ || RB

⇒ PQ || AB [∵ RB is a part of AB]

⇒ PQ || AB and DC [∵ AB || DC (Given)]

This proves (i).

Again, P and Q are the mid-points of sides DR and DB respectively in ∆DRB.

∴ PQ = 1/2 RB ⇒ PQ = 1/2 (AB – AR)

⇒ PQ = 1/2 (AB – DC) [From (ii), AR = DC]

This proves (ii).

**Example 25: **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.

**Solution:** Join AC, meeting BD at O.

Since the diagonals of a parallelogram bisect each other, we have OA = OC and OD = OB.

Now, OD = OB and DX = BY

⇒ OD – DX = OB – BY ⇒ OX = OY

Now, OA = OC and OX = OY

∴ CXAY is a quadrilateral whose diagonals bisect each other.

∴ CXAY is a ||gm

**Example 26: **Prove that the four triangles formed by joining in pairs, the mid-points of three sides of a triangle, are concurrent to each other.

**Solution:** **Given:** A triangle ABC and D,E,F are the mid-points of sides BC, CA and AB respectively.

**To Prove:**

∆ AFE ≅ ∆FBD ≅ ∆EDC ≅ ∆DEF.

**Proof:** Since the segment joining the mid-points of the sides of a triangle is half of the third side. Therefore,

DE = 1/2 ⇒ DE = AF = BF …. (i)

EF = 1/2 ⇒ EF = BD = CD …. (ii)

DF = 1/2 ⇒ DF = AE = EC ….(iii)

Now, in ∆s DEF and AFE, we have

DE = AF [From (i)]

DF = AE [From (ii)]

and, EF = FE [Common]

So, by SSS criterion of congruence,

∆ DEF ≅ ∆AFE

Similarly, ∆DEF ≅ ∆FBD and ≅ DEF ≅ ∆EDC

Hence, ∆ AFE ≅ ∆FBD ≅ ∆EDC ≅ ∆DEF.

**Example 27: **In fig, AD is the median and DE || AB. Prove that BE is the median.

**Solution:** In order to prove that BE is the median, it is sufficient to show that E is the mid-point of AC.

Now, AD is the median in ∆ABC

⇒ D is the mid-point of BC.

Since DE is a line drawn through the mid-point of side BC of ∆ABC and is parallel to AB (given). Therefore, E is the mid-point of AC. Hence, BE is the median of ∆ABC.

**Example 28: **Let ABC be an isosceles triangle with AB = AC and let D,E,F be the mid-points of BC, CA and AB respectively. Show that

AD ⊥ FE and AD is bisected by FE.

**Solution: Given:** 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.

**To Prove:** AD ⊥ FE and AD is bisected by FE.

**Constructon:** Join DE and DF.

**Proof:** Since the segment joining the mid-points of two sides of a triangle is parallel to third side and is half of it. Therefore,

DE || AB and DE = 1/2 AB

Also, DF || AC and DF = 1/2 AC

But, 1/2 AB = 1/2 AC [Given]

⇒ AB = AC

⇒ DE = DF …. (i)

Now, DE = 1/2 AB ⇒ DE = AF …. (ii)

and, DF = 1/2 AC ⇒ DF = AE …(iii)

From (i), (ii) and (iii) we have

DE = AE = AF = DF

⇒ DEAF is a rhombus.

⇒ Diagonals AD and FE bisect each other at right angle.

AD ⊥ FE and AD is bisected by FE.

**Example 29: **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.

**Solution:** ABCD is a parallelogram.

⇒ AD = BC and AD || BC

⇒ 1/3 AD = 1/3 BC and AD || BC

⇒ AP = CQ and AP || CQ

Thus, APCQ is a quadrilateral such that one pair of opposite side AP and CQ are parallel and equal.

Hence, APCQ is a parallelogram.

**Example 30: **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.

**Solution: ** Since 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.

**Example 31: **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

**Solution:** **Given:** A ∆ABC with P,Q and R as the mid-points of BC, CA and AB respectively. PR and BQ meet at X and CR and PQ meet at Y.

**Construction:** Join “X and Y.

**Proof:** 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.

Similarly, Y is the mid-point of PQ.

Now, consider ∆PQR. XY is the line segment joining the mid-points of sides PR and PQ.

∴ XY = 1/2 RQ …. (i)

But RQ = 1/2 BC [From (i)]

Hence, XY = 1/4 BC

**Example 32: **Show that the quadrilateral, formed by joining the mid-points of the sides of a square, is also a square.

**Solution:** **Given:** 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.

**To Prove:** PQRS is a square.

**Construction:** Join AC and BD.

Proof : In ∆ABC, P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ = 1/2 AC …. (i)

In ∆ADC, R and S are the mid-points of CD and AD respectively.

∴ RS || AC and RS = 1/2 AC ….(ii)

From (i) and (ii), we have

PQ || RS and PQ = RS ….(iii)

Thus, in quadrilateral PQRS one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram.

Now, in ∆s PBQ and RCQ, we have

PB = RC

Now, PQ || AC [From (i)]

⇒ PM || NO ….(vi)

Since P and S are the mid-points of AB and AD respectively.

PS || BD

⇒ PM || MO ….(vii)

Thus, in quadrilateral PMON, we have

PM || NO [From (vi)]

PN || MO [From (vii)]

So, PMON is a parallelogram.

⇒ ∠MPN = ∠MON

⇒ ∠MPN = ∠BOA [∵ ∠MON = ∠BOA]

⇒ ∠MPN = 90°

⇒ ∠QPS = 90°

Thus, PQRS is a quadrilateral such that

PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square.

**Example 33: **∆ABC is a triangle right angled at B ; and P is the mid-point of AC. Prove that PB = PA = 1/2 AC.

**Solution:** Given : ∆ABC right angled at B, P is the mid-point of AC.

**To Prove:** PB = PA = 1/2 AC.

**Construction:** Through P draw PQ || BC meeting AB at Q.

**Proof:** Since PQ || BC. Therefore,

∠AQP = ∠ABC [Corresponding angles]

⇒ ∠AQP = 90° [∵ ∠ABC = 90°]

But, ∠AQP + ∠BQP = 180°

[∵ ∠AQP & ∠BQP are angles of a linear pair]

∴ 90° + ∠BQP = 180°

⇒ ∠BQP = 90°

Thus, ∠AQP = ∠BQP = 90°

Now, in ∆ABC, P is the mid-point of AC and

PQ || BC. Therefore, Q is the mid-point of AB i.e, AQ = BQ.

Consider now ∆s APQ and BPQ.

we have, AQ = BC [Proved above]

∠AQP = ∠BQP [From (i)]

and, PQ = PQ

So, by SAS cirterion of congruence

∆APQ ≅ ∠BPQ

⇒ PA = PB

Also,

PS = 1/2 AC, since P is the mid-point of AC

Hence, PA = PB = 1/2 AC.

**Example 34: **Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.

**Solution:** **Given:** 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.

**To Prove:** PQRS is rhombus.

**Construction:** Join AC.

**Proof:** In ∆ABC, P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ = AC …. (i)

In ∆ ADC, R and S are the mid-points of CD and AD respectively.

∴ SR || AC and SR = 1/2 AC …. (ii)

From (i) and (ii), we get

PQ || SR and PQ = SR ….(iii)

⇒ PQRS is a parallelogram.

Now, ABCD is a rectangle.

⇒ AD = BC ⇒ 1/2 AD = 1/2 BC

⇒ AS = BQ ….(iv)

In ∆s APS and BPQ , we have

AP = BP [∴P is the mid-point of AB]

∠PAS = ∠PBQ [Each equal to 90°]

and, AS = BQ [From (iv)]

So, by SAS criterion of congruence

∆APS ≅ ∆BPQ

PS = PQ ….(v)

[∵ Corresponding parts of congruent triangles are equal]

From (iii) and (v), we obtain that PQRS is a parallelogram such that PS = PQ i.e., two adjacent sides are equal.

Hence, PQRS is a rhombus.

Asritha says

Wow thanks for the solutions.

ashay says

It just amazing