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In Fig. 6.63, D is a point on side BC of Delta ABC such that BD by CD equals AB by AC Prove that AD is the bisector of angle BAC.

Q9   In Fig. 6.63, D is a point on side BC of D ABC such that   \frac{BD }{CD} = \frac{AB}{AC}
        Prove that AD is the bisector of \angle BAC.

      

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Produce BA to P, such that AP=AC and join P to C.

\frac{BD }{CD} = \frac{AB}{AC}         (Given )

\Rightarrow \frac{BD }{CD} = \frac{AP}{AC}

Using converse of Thales theorem,

AD||PC \Rightarrow \angle BAD=\angle APC............1(Corresponding angles)

\Rightarrow \angle DAC=\angle ACP............2           (Alternate angles)

By construction,

AP=AC

\Rightarrow \angle APC=\angle ACP............3

From equation 1,2,3, we get

\Rightarrow \angle BAD=\angle APC

Thus, AD bisects angle BAC.

 

 

 

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