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  • D is a point on the side BC of a \triangleABC such that AD bisects \angleBAC. Then (A) BD = CD(B) BA > BD(C) BD > BA(D) CD > CA

$D$ is a point on the side $B C$ of a $\triangle A B C$ such that $A D$ bisects
$\angle B A C$. Then
(A) $B D=C D$
(B) $B A>B D$
(C) $B D>B A$
(D) $C D>C A$

Answers (1)

Solution.

$B C$ is the side of $\triangle A B C$, and $D$ is the point on the side of $B C$.

$\begin{aligned} & \text { In } \triangle A B C, \\ & \angle B A D=\angle C A D \quad \text { (Given that } A D \text { bisects } B A C) \\ & \text { In } \triangle A D C, \\ & \angle B D A>\angle B A D \quad(\because \text { Exterior angle of triangle is greater than interior opposite angle) } \\ & \angle B A D=\angle C A D \\ & \text { Then } B A>B D \quad(\because \text { Side opposite to greater angle is longer) }\end{aligned}$

Hence option (A) is correct.

Posted by

infoexpert24

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