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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 BC of a \triangleABC such that AD bisects \angleBAC. Then

(A) BD = CD

(B) BA > BD

(C) BD > BA

(D) CD > CA

Answers (1)

            [A]

Solution.        

BC is the side of \triangleABC, D is the point on side BC.

In \triangleABC,

\angleBAD = \angleCAD         (Given that AD bisects ÐBAC)

In \triangleADC,

\angleBDA > \angleBAD        (\because Exterior angle of triangle is greater than interior opposite angle)

\angleBAD = \angleCAD

Then BA > BD                        (\because Side opposite to greater angle is longer)

Hence option (A) is correct.

Posted by

infoexpert24

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