Question

2. In the given figure ABC and DBC are two isowcolas triangles on the same base BC.
Show that LABD-LACB
A
kec
D​

Answer 1

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

Answer 2

Question :

In the given figure ABC and DBC are two isosceles triangles on the same base BC.

Show that ∠ABD = ∠ACD.

Given :

ABC and DBC are two isosceles Δ on same base BC.

To prove :

∠ABD = ∠ACD

Solution :

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.

Thank you.