Given: ∆ABC in which ∠B = ∠C To Prove: AB = AC Construction: We draw the bisector of ∠A which meets BC in D. Proof: In ∆ABD and ∆ACD we have ∠B = ∠C [Given] ∠BAD = ∠CAD [∵AD is bisector of ∠A] AD = AD [Common side] ∴ By AAS criterion of congruence, we get ∆ABD ≅ ∆ACD ⇒ AB = AC MARK IT AS BRAINLIEST Reply
Given: ∆ABC in which ∠B = ∠C
To Prove: AB = AC
Construction: We draw the bisector of ∠A which meets BC in D.
Proof: In ∆ABD and ∆ACD we have ∠B = ∠C
[Given] ∠BAD = ∠CAD [∵AD is bisector of ∠A]
AD = AD [Common side]
∴ By AAS criterion of congruence, we get ∆ABD ≅
∆ACD ⇒ AB = AC
MARK IT AS BRAINLIEST