Question

show that in an isosceles triangle, Angle opposites to equal sides are equal.​

Answer 1

Step-by-step explanation:

Let us draw the bisector of ∠A

Let D be the point of intersection of this bisector of ∠A and BC.

Therefore ,by construction ∠BAD = ∠CAD.

In ∆BAD and ∆DAC,

AB = AC (Given)

∠BAD = ∠CAD (By construction)

AD = AD (Common side in both triangle)

So, ∆BAD ≅ ∆CAD (By SAS rule)

So, ∠ABD = ∠ACD, since they are corresponding angles of congruent triangles.

So, ∠B = ∠C

Hence, Proved that an angle opposite to equal sides of an isosceles triangle is equal.

Answer 2

Step-by-step explanation:

Let △ABC be an isosceles triangle such that AB =AC Then we have to prove that ∠B=∠C Draw the bisector AD of ∠A meeting BC in D

Now in triangles ABD and ACD We have AB=AC (Given)

∠BAD=∠CAD (because AD is bisector of ∠A

AD=AD (Common side)

Therefore by SAS congruence condition we have

△ABC≅△ACD

⇒∠B=∠C

(Corresponding parts of congruent triangles are equal )

hope this helps u