Question

plz anyone tell me answer fast​

Answer 1

Here in ∆ADB and ∆ADC.

i) Three pair of equal parts are:

AD = AD ( common side )

BD = CD ( as d is the mid point of BC )

AB = AC ( given in the question )

ii) Now,

by SSS Congruency rule,

∆ADB ≈ ∆ADC

iii) As both triangles are congruent to each other we can compare them and say

angelB = angleC.