Question

Prove that angles opposite to two equal sides of a triangle are equal.

Answer 1

Given: ΔABC such that AB = AC
To Prove: ∠ABC = ∠ACB
Construction: Draw AD, the bisector of ∠BAC to meet BC at D.

Proof: In Δ ADB and Δ ADC,
AB = AC ... (Given)
AD = AD ... (Common)
∠BAD = ∠CAD ... (by construction)
∴ Δ ABD ≅ Δ ACD ... (SAS Cong. Axiom)
∴ ∠ABD = ∠ACD (CPCT)
⇒ ∠ABC = ∠ACB

Answer 2

Heya ...

Here the ans is.._____

Given : a triangle ABC in which AB =AC
RTP : angle ABC =angle ACB
Construction : draw a bisector AD of A so as to intersect BC at D
Proof : in triangle ABC and ADC,
AD=AD Common
AB =AC given
angle BAD =angle CAD

by construction
triangle ABC is congruent to triangle ADC
therefore by SAS congruence rule they r equal

Hope my ans helps u dear..☺

Plz...Make my ans as brainliest if I deserve. ❤

@shuzu