Question

2) In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of
triangle such that angleBCD = angleCBD. Prove that AD bisects angleBAC of triangleABC
(marks-2)​

Answer 1

Given, ∆ABC where AB = AC, D is any mid point in the interior of triangle Such that ∠DBC = ∠DCB.

To Prove : AD bisects ∠BAC

Proof ⇒    

            Join AD

Since, ∠DBC = ∠DCB

          ∴ DB = DC

Now, In ∆ABD and ∆ACD,

            AB = AC (Given)

            AD = DA (Common)

            DB = DC  (proved above)

 ∴ ∆ABD ≅ ∆ACD (by SSS)

     ∠BAD = ∠CAD (by CPCT)

Hence, AD bisects ∠BAC.