{"id":11381,"date":"2022-07-08T01:17:20","date_gmt":"2022-07-07T19:47:20","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11381"},"modified":"2022-07-08T01:17:23","modified_gmt":"2022-07-07T19:47:23","slug":"if-two-sides-and-a-median-bisecting-the-third-side-of-a-triangle-are-respectively-proportional-to-the-corresponding-side-and-the-median-of-another-triangle-the-prove-that-the-two-triangles-are-simila","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-two-sides-and-a-median-bisecting-the-third-side-of-a-triangle-are-respectively-proportional-to-the-corresponding-side-and-the-median-of-another-triangle-the-prove-that-the-two-triangles-are-simila\/","title":{"rendered":"If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding side and the median of another triangle, the prove that the two triangles are similar."},"content":{"rendered":"
Given<\/strong> : Two triangles ABC and DEF, in which AP and DQ are the medians, such that<\/p>\n \\(AB\\over DE\\) = \\(AC\\over DF\\) = \\(AP\\over DQ\\)<\/p>\n To prove<\/strong> : \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF<\/p>\n Construction<\/strong> : Produce line AP to G, so that PG = AP. Join CG. And Produce line DQ to H, so that QH = DQ. Join FH.<\/p>\n Proof<\/strong> : In triangles APB and GPC, BP = CP\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (because AP is the median)<\/p>\n AP = GP\u00a0 \u00a0 \u00a0 \u00a0 (by construction)<\/p>\n and\u00a0 \u00a0 \\(\\angle\\) APB = \\\\(\\angle\\) CPG\u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles)<\/p>\n \\(\\therefore\\)\u00a0 By SAS theorem of congruence,<\/p>\n \\(\\triangle\\) APB \\(\\cong\\) \\(\\triangle\\) GPC<\/p>\n \\(\\implies\\)\u00a0 AB = GC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(C.P.C.T)<\/p>\n Again, in triangles DQE and HQF,<\/p>\n EQ = FQ\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(because DQ is the median)<\/p>\n DQ = HQ\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (by construction)<\/p>\n and\u00a0 \\(\\angle\\) DQE = \\(\\angle\\) HQF\u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles)<\/p>\n \\(\\therefore\\)\u00a0 By SAS theorem of congruence,<\/p>\n \\(\\triangle\\) DQE \\(\\cong\\) \\(\\triangle\\) HQF<\/p>\n \\(\\implies\\)\u00a0 DE = HF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………(2)\u00a0 \u00a0 \u00a0 \u00a0 (C.P.C.T)<\/p>\n Now,\u00a0 \\(AB\\over DE\\) = \\(AC\\over DF\\) = \\(AP\\over DQ\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (given)<\/p>\n From (1) and (2),\u00a0 \u00a0 \u00a0AB = GC\u00a0 and\u00a0 \u00a0DE = HF<\/p>\n \\(\\implies\\)\u00a0 \\(GC\\over HF\\) = \\(AC\\over DF\\) = \\(AP\\over DQ\\)<\/p>\n \\(\\implies\\) \\(GC\\over HF\\) = \\(AC\\over DF\\) = \\(2AP\\over 2DQ\\)<\/p>\n \\(\\implies\\)\u00a0 \\(GC\\over HF\\) = \\(AC\\over DF\\) = \\(AG\\over DH\\)<\/p>\n By SSS similarity,<\/p>\n \\(\\triangle\\) AGC ~ \\(\\triangle\\) DHF<\/p>\n \\(\\implies\\)\u00a0 \\(\\angle\\) 1 = \\(\\angle\\) 2<\/p>\n Similarly, \\(\\angle\\) 3 = \\(\\angle\\) 4<\/p>\n Thus,\u00a0 \\(\\angle\\) 1 + \\(\\angle\\) 3 = \\(\\angle\\) 2 + \\(\\angle\\) 4<\/p>\n \\(\\implies\\)\u00a0 \\(\\angle\\) A = \\(\\angle\\) D\u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(3)<\/p>\n Thus, in triangles ABC and DEF,<\/p>\n \\(\\angle\\) A = \\(\\angle\\) D\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(from 3)<\/p>\n and \\(AB\\over DE\\) = \\(AC\\over DF\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0(given)<\/p>\n So, By SAS similarity,<\/p>\n \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : Two triangles ABC and DEF, in which AP and DQ are the medians, such that \\(AB\\over DE\\) = \\(AC\\over DF\\) = \\(AP\\over DQ\\) To prove : \\(\\triangle\\) ABC ~ \\(\\triangle\\) DEF Construction : Produce line AP to G, so that PG = AP. Join CG. And Produce line DQ to H, …<\/p>\n