{"id":11558,"date":"2022-07-18T14:50:15","date_gmt":"2022-07-18T09:20:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11558"},"modified":"2022-07-18T14:50:19","modified_gmt":"2022-07-18T09:20:19","slug":"in-the-figure-two-chords-ab-and-cd-intersect-each-other-at-the-point-p-prove-that-i-triangle-apc-triangle-dpb-ii-ap-pb-cp-dp","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-two-chords-ab-and-cd-intersect-each-other-at-the-point-p-prove-that-i-triangle-apc-triangle-dpb-ii-ap-pb-cp-dp\/","title":{"rendered":"In the figure, two chords AB and CD intersect each other at the point P. Prove that (i) \\(\\triangle\\) APC ~ \\(\\triangle\\) DPB (ii) AP.PB = CP.DP"},"content":{"rendered":"
(i)<\/strong>\u00a0 In \\(\\triangle\\)s PAC and PDB, we have :<\/p>\n \\(\\angle\\) APC = \\(\\angle\\) DPB\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(vertically opp. angles)<\/p>\n \\(\\angle\\) CAP = \\(\\angle\\) BDP\u00a0 \u00a0 \u00a0 \u00a0 (angles in same segment of circle are equal)<\/p>\n \\(\\therefore\\)\u00a0 By AA similarity, we have :<\/p>\n \\(\\triangle\\) APC ~ \\(\\triangle\\) DPB\u00a0<\/strong><\/p>\n (ii)<\/strong>\u00a0 Since \\(\\triangle\\)s APC ~ DPB, therefore<\/p>\n \\(AP\\over DP\\) = \\(CP\\over PB\\)\u00a0 or\u00a0 PA.PB = CP.DP.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : (i)\u00a0 In \\(\\triangle\\)s PAC and PDB, we have : \\(\\angle\\) APC = \\(\\angle\\) DPB\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(vertically opp. angles) \\(\\angle\\) CAP = \\(\\angle\\) BDP\u00a0 \u00a0 \u00a0 \u00a0 (angles in same segment of circle are equal) \\(\\therefore\\)\u00a0 By AA similarity, we have : \\(\\triangle\\) APC ~ \\(\\triangle\\) DPB\u00a0 (ii)\u00a0 Since \\(\\triangle\\)s APC ~ …<\/p>\n