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