{"id":11394,"date":"2022-07-08T01:22:44","date_gmt":"2022-07-07T19:52:44","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11394"},"modified":"2022-07-08T01:22:45","modified_gmt":"2022-07-07T19:52:45","slug":"if-ad-and-pm-are-medians-of-triangles-abc-and-pqr-respectively-where-triangle-abc-triangle-pqr-prove-that-abover-pq-adover-pm","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-ad-and-pm-are-medians-of-triangles-abc-and-pqr-respectively-where-triangle-abc-triangle-pqr-prove-that-abover-pq-adover-pm\/","title":{"rendered":"If AD and PM are medians of triangles ABC and PQR respectively, where \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR, prove that \\(AB\\over PQ\\) = \\(AD\\over PM\\)."},"content":{"rendered":"
Given<\/strong> : AD and PM are the medians of triangles ABC and PQR respectively, where \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR.<\/p>\n To Prove<\/strong> : \\(AB\\over PQ\\) = \\(AD\\over PM\\)<\/p>\n Proof<\/strong>\u00a0 : In triangles ABD and PQM, we have<\/p>\n \\(\\angle\\) B = \\(\\angle\\) D\u00a0 \u00a0 \u00a0 \u00a0 (because \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR)<\/p>\n Since AD and PM are the medians to BC and QR respectively and \\(AB\\over PQ\\) = \\(BC\\over QR\\)<\/p>\n \\(AB\\over PQ\\) = \\({1\\over2}BC\\over {1\\over 2}QR\\)<\/p>\n So, \\(AB\\over PQ\\) = \\(BD\\over QM\\)<\/p>\n \\(\\therefore\\)\u00a0 By SAS similarity,<\/p>\n \\(\\triangle\\) ABD ~ \\(\\triangle\\) PQM<\/p>\n So, \\(AB\\over PQ\\) = \\(BD\\over QM\\) = \\(AD\\over PM\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : AD and PM are the medians of triangles ABC and PQR respectively, where \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR. To Prove : \\(AB\\over PQ\\) = \\(AD\\over PM\\) Proof\u00a0 : In triangles ABD and PQM, we have \\(\\angle\\) B = \\(\\angle\\) D\u00a0 \u00a0 \u00a0 \u00a0 (because \\(\\triangle\\) ABC ~ \\(\\triangle\\) PQR) Since AD …<\/p>\n