RS Aggarwal Class 9 Solutions Quadrilaterals and Parallelograms<\/a><\/li>\n<\/ul>\nProperties of Cyclic Quadrilaterals Example Problems With Solutions<\/h2>\n Example 1: \u00a0 \u00a0<\/strong>Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic. \nSolution:<\/strong> \n \nGiven a quadrilateral ABCD with internal angle bisectors AF, BH, CH and DF of angles A, B, C and D respectively and the points E, F, G and H form a quadrilateral EFGH. \nTo prove that EFGH is a cyclic quadrilateral. \n\u2220HEF = \u2220AEB [Vertically opposite angles] ——– (1) \nConsider triangle AEB, \n\u2220AEB + \\(\\frac { 1 }{ 2 } \\) \u2220A + \\(\\frac { 1 }{ 2 } \\) \u2220 B = 180\u00b0 \n\u2220AEB = 180\u00b0 \u2013 \\(\\frac { 1 }{ 2 } \\) (\u2220A + \u2220 B) ——– (2) \nFrom (1) and (2), \n\u2220HEF = 180\u00b0 \u2013 \\(\\frac { 1 }{ 2 } \\) (\u2220A + \u2220 B) ——— (3) \nSimilarly, \u2220HGF = 180\u00b0 \u2013 \\(\\frac { 1 }{ 2 } \\) (\u2220C + \u2220 D) ——– (4) \nFrom 3 and 4, \n\u2220HEF + \u2220HGF = 360\u00b0 \u2013 \\(\\frac { 1 }{ 2 } \\) (\u2220A + \u2220B + \u2220C + \u2220 D) \n= 360\u00b0 \u2013 \\(\\frac { 1 }{ 2 } \\) (360\u00b0) \n= 360\u00b0 \u2013 180\u00b0 \n= 180\u00b0 \nSo, EFGH is a cyclic quadrilateral since the sum of the opposite angles of the quadrilateral is 180\u00b0<\/p>\nExample 2: \u00a0 \u00a0<\/strong>ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \u2220DBC = 70\u00ba, \u2220BAC is 30\u00ba, Find \u2220BCD. Further, if AB = BC, find \u2220ECD. \nSolution:<\/strong> \n \nFor chord CD, \n\u2220CBD = \u2220CAD … Angles in same segment \n\u2220CAD = 70\u00b0 \n\u2220BAD = \u2220BAC + \u2220CAD = 30\u00b0 + 70\u00b0 = 100\u00b0 \n\u2220BCD + \u2220BAD = 180\u00b0 …Opposite angles of a cyclic quadrilateral \n\u21d2 \u2220BCD + 100\u00b0 = 180\u00b0 \n\u21d2 \u2220BCD = 80\u00b0 \nIn \u25b3ABC \nAB = BC (given) \n\u2220BCA = \u2220CAB … Angles opposite to equal sides of a triangle \n\u2220BCA = 30\u00b0 \nAlso, \u2220BCD = 80\u00b0 \n\u2220BCA + \u2220ACD = 80\u00b0 \n\u21d2 30\u00b0 + \u2220ACD = 80\u00b0 \n\u2220ACD = 50\u00b0 \n\u2220ECD = 50\u00b0<\/p>\nExample 3: \u00a0 \u00a0<\/strong>Prove that a cyclic parallelogram is a rectangle. \nSolution:<\/strong> \n \nGiven that, ABCD is a cyclic parallelogram. \nTo prove, ABCD is a rectangle. \nProof: \n\u22201 + \u22202 = 180\u00b0 …Opposite angles of a cyclic parallelogram \nAlso, Opposite angles of a cyclic parallelogram are equal. \nThus, \n\u22201 = \u22202 \n\u21d2 \u22201 + \u22201 = 180\u00b0 \n\u21d2 \u22201 = 90\u00b0 \nOne of the interior angle of the parallelogram is right angled. Thus, ABCD is a rectangle.<\/p>\n","protected":false},"excerpt":{"rendered":"What are the Properties of Cyclic Quadrilaterals? Cyclic quadrilateral If all four points of a quadrilateral are on circle then it is called cyclic Quadrilateral. A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices P, Q, R and S. Let a cyclic quadrilateral be […]<\/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":[248,1277,1280,1278,1279],"yoast_head":"\n
What are the Properties of Cyclic Quadrilaterals? - A Plus Topper<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n