{"id":11347,"date":"2022-07-07T22:46:39","date_gmt":"2022-07-07T17:16:39","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11347"},"modified":"2022-07-07T22:46:43","modified_gmt":"2022-07-07T17:16:43","slug":"in-the-figure-if-triangle-abe-cong-triangle-acd-show-that-triangle-ade-triangle-abc","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-if-triangle-abe-cong-triangle-acd-show-that-triangle-ade-triangle-abc\/","title":{"rendered":"In the figure, if \\(\\triangle\\) ABE \\(\\cong\\) \\(\\triangle\\) ACD, show that \\(\\triangle\\) ADE ~ \\(\\triangle\\) ABC."},"content":{"rendered":"
It is given that \\(\\triangle\\) ABE \\(\\cong\\) \\(\\triangle\\) ACD<\/p>\n
\\(\\therefore\\)\u00a0 \u00a0AB = AC<\/p>\n
and\u00a0 AE = AD<\/p>\n
[ because corresponding parts of congruent triangles are equal ]<\/p>\n
So,\u00a0 \\(AB\\over AD\\) = \\(AC\\over AE\\)\u00a0 \u00a0 or\u00a0 \u00a0 \u00a0\\(AB\\over AC\\) = \\(AD\\over AE\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(1)<\/p>\n
\\(\\therefore\\)\u00a0 In triangles ADE and ABC, we have :<\/p>\n
\\(AB\\over AC\\) = \\(AD\\over AE\\)<\/p>\n
and\u00a0 \\(\\angle\\) BAC = \\(\\angle\\) DAE<\/p>\n
Thus, by SAS criterion of similarity, \\(\\triangle\\) ADE ~ \\(\\triangle\\) ABC.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : It is given that \\(\\triangle\\) ABE \\(\\cong\\) \\(\\triangle\\) ACD \\(\\therefore\\)\u00a0 \u00a0AB = AC and\u00a0 AE = AD [ because corresponding parts of congruent triangles are equal ] So,\u00a0 \\(AB\\over AD\\) = \\(AC\\over AE\\)\u00a0 \u00a0 or\u00a0 \u00a0 \u00a0\\(AB\\over AC\\) = \\(AD\\over AE\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………(1) \\(\\therefore\\)\u00a0 In triangles ADE and ABC, …<\/p>\n