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  • In given figure, AD is the bisector of angle BAC. Prove that AB greater than BD.

In given figure, AD is the bisector of \angleBAC. Prove that AB > BD.

Answers (1)

Given: In \triangleABC, AD is bisector of \angleBAC.

To prove: AB > BD

Proof: We know that

Exterior angle of triangle is greater than each of opposite interior angles then

In \triangleADC

\angleADB > \angleCAD                     (From above result)

Now, \angleBAD = \angleDAC            (\becauseAD is bisector of \angleBAC)

\Rightarrow \angleADB > \angleBAD

\therefore AB > BD                             (Since, side opposite to greater angle is longer)

Hence proved.

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