Answer:
Answer:
Solution given;
<BCA=<ACD
<ABC=<ADC
now
In traingle ABC & ∆ ACD
AC=AC[common base]
<BCA=<ACD[bisector bisect it]
<ABC=<ADC[Given]
∆ABC is congruent to ∆ ACD by S.A.A. axiom
AB is congruent to AD.
[ corresponding sides of a congruent triangle are equal]
Hence proved.
