Question

In fig., ABC and DBC are two isosceles triangle on the same base BC such that AB AC and DB = DC. Prove that angle ABD = angle ACD.​

Answer 1

$\bf \huge \underline {\underline{ANSWER}}$

Given ΔABC and ΔDBC are two isosceles triangles on the same base BC

In isosceles ΔABC

AB=AC

Then ∠ABC=∠ACB....(1)

In isosceles ΔBDC

BD=DC

Then ∠CBD=∠BCD....(2)

Add (1) and (2) we get

∠ABC+∠CBD=∠ACB+∠BCD

But ΔABC and ΔDBC on same base

∴∠ABD=∠ACD