Question

In the given figure, BD=CE and BE=CD. Prove that ABC is an isosceles triangle.​

Answer 1

We will prove this using congruency and linear pairs.

In △DBC and △ECB,

DC = EB (Given)

BD = CE (Given)

BC = BC (Common)

So, △DBC ≅ △ECB by SSS congruence criterion

Now,

∠DBC = ∠ECB ..(1)                by CPCT(Common Parts of Congruent Triangles)

Now, ∠DBC and ∠ABC form a linear pair

So, ∠DBC + ∠ABC = 180

Therefore, ∠ABC = 180 - ∠DBC

Also,

∠ECB and ∠ACB form a linear pair

So, ∠ECB + ∠ACB = 180

Therefore, ∠ACB = 180 - ∠ECB

But we know that ∠ECB = ∠DBC from (1)

So,

∠ABC = 180 - ∠DBC = 180 - ∠ECB

And,

∠ACB = 180 - ∠ECB = 180 - ∠DBC

Therefore,

∠ABC = ∠ACB

Now, by Theorom 2 of the Isosceles Triangle Property, we know that sides opposite to equal angles in a triangle are equal

Therefore, BA = AC

Therefore, △ABC is an isosceles triangle

Hope this helped! This took quite some time to type out so would appreciate if you mark it as brainliest :D