{"id":11563,"date":"2022-07-18T14:57:33","date_gmt":"2022-07-18T09:27:33","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11563"},"modified":"2022-07-18T14:57:37","modified_gmt":"2022-07-18T09:27:37","slug":"in-the-figure-d-is-a-point-on-side-bc-of-triangle-abc-such-that-bdover-cd-abover-ac-prove-that-ad-is-the-bisector-of-angle-bac","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-d-is-a-point-on-side-bc-of-triangle-abc-such-that-bdover-cd-abover-ac-prove-that-ad-is-the-bisector-of-angle-bac\/","title":{"rendered":"In the figure, D is a point on side BC of triangle ABC such that \\(BD\\over CD\\) = \\(AB\\over AC\\). Prove that AD is the bisector of \\(\\angle\\) BAC."},"content":{"rendered":"
Given<\/strong> : ABC is a triangle and D is point on BC such that \\(BD\\over CD\\) = \\(AB\\over AC\\)<\/p>\n To Prove<\/strong> : AD is the bisector of \\(\\angle\\) BAC.<\/p>\n Construction<\/strong> : Produce line BA to E such that line AE = AC. Join CE.<\/p>\n Proof<\/strong> : In \\(\\triangle\\) AEC, since AE = AC, hence<\/p>\n \\(\\angle\\) AEC = \\(\\angle\\) ACE\u00a0 \u00a0 \u00a0 (angles opposite to equal sides of triangle are equal)<\/p>\n Now,\u00a0 \\(BD\\over CD\\) = \\(AB\\over AC\\)\u00a0 \u00a0 \u00a0(given)<\/p>\n So,\u00a0 \\(BD\\over CD\\) = \\(AB\\over AE\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (AE = AC, by construction)<\/p>\n \\(\\therefore\\)\u00a0 By converse of Basic Proportionality theorem(Thales Theorem),<\/p>\n DA || CE<\/p>\n Now, Since CA is a traversal, we have :<\/p>\n \\(\\angle\\) BAD = \\(\\angle\\) AEC\u00a0 \u00a0 \u00a0 ……..(2)\u00a0 \u00a0 \u00a0 \u00a0 [corresponding angles]<\/p>\n and\u00a0 \\(\\angle\\) DAC = \\(\\angle\\) ACE\u00a0 \u00a0 \u00a0 \u00a0……..(3)\u00a0 \u00a0 \u00a0 (alternate angles)<\/p>\n Also,\u00a0 \\(\\angle\\) AEC = \\(\\angle\\) ACE\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(from 1)<\/p>\n From (2) and (3),<\/p>\n \\(\\angle\\) BAD = \\(\\angle\\) DAC<\/p>\n Thus,\u00a0 AD bisects \\(\\angle\\) BAC.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : ABC is a triangle and D is point on BC such that \\(BD\\over CD\\) = \\(AB\\over AC\\) To Prove : AD is the bisector of \\(\\angle\\) BAC. Construction : Produce line BA to E such that line AE = AC. Join CE. Proof : In \\(\\triangle\\) AEC, since AE = AC, …<\/p>\n