Question

Hi guys!!

How to solve this question...

Pls help me....

Answer 1

$\textbf{ Hello Mate! }$

Given : AB = AC and < ABO = < ACO

To Prove : triangles ABO =~ ACO

Proof : AB = AC

Angle opposite to equal sides are equal.

Hence, < AOB = < AOC ____(1)

In triangle AOB and AOC

< AOB = < AOC [ From (1) ]

< ABO = < ACO [ Given ]

AD = DA [ Common ]

Hence triangles AOB =~ AOC by AAS congruency.

$\textbf{ Q.E.D }$

Have great future ahead!