Question

in the given figure ABC is an isosceles triangle in which is equal to AC also D is a point such that BD equal to CD prove that a d bisect angle A and angle d

Answer 1

In ∆ADB & ∆ADC
AB=AC (given)
AD=AD (common)
angle ADB=angle ADC (each 90 degree)
∆ADB is congurant to ∆ADC (By R.H.S rule)