{"id":11291,"date":"2022-07-01T01:17:32","date_gmt":"2022-06-30T19:47:32","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11291"},"modified":"2022-07-01T01:18:02","modified_gmt":"2022-06-30T19:48:02","slug":"in-the-figure-a-b-and-c-are-points-on-op-oq-and-or-respectively-such-that-ab-pq-and-ac-pr-show-that-bc-qr","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-a-b-and-c-are-points-on-op-oq-and-or-respectively-such-that-ab-pq-and-ac-pr-show-that-bc-qr\/","title":{"rendered":"In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR."},"content":{"rendered":"
Given<\/strong> : O\u00a0 is any point within triangle PQR, AB || PQ and AC || PR<\/p>\n To Prove<\/strong> :\u00a0 BC || QR<\/p>\n Construction<\/strong> : Join BC<\/p>\n Proof<\/strong> : In triangle OPQ,<\/p>\n Given,\u00a0 \u00a0 \u00a0 AB || PQ<\/p>\n By basic proportionality theorem,<\/p>\n \\(OA\\over AP\\) = \\(OB\\over BQ\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n In triangle OPR,<\/p>\n Given,\u00a0 \u00a0 \u00a0 \u00a0 AC || PR<\/p>\n By basic proportionality theorem,<\/p>\n \\(OA\\over AP\\) = \\(OC\\over CR\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(2)<\/p>\n From (1) and (2), we obtain that<\/p>\n \\(OB\\over BQ\\) = \\(OC\\over CR\\)<\/p>\n Hence, by converse of basic proportionality theorem, we have<\/p>\n BC || QR<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : O\u00a0 is any point within triangle PQR, AB || PQ and AC || PR To Prove :\u00a0 BC || QR Construction : Join BC Proof : In triangle OPQ, Given,\u00a0 \u00a0 \u00a0 AB || PQ By basic proportionality theorem, \\(OA\\over AP\\) = \\(OB\\over BQ\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 …<\/p>\n