Question

in figure ABC is an isosceles triangle with ab is equal to AC also D is a point such that BD is equal to CD prove that ad bisects angle A and angle d

Answer 1

In ∆ABD and ∆ACD, we have
AB = AC (Given)
∠B = ∠C (Angles opposite to equal sides)
AD = AD (Common)
∆ABD ≅ ∆ACD
Thus, their corresponding parts are equal.
∠BAD = ∠CAD
So, AD bisects ∠A

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