Question

prove that angle opposite to equal sides of isosceles triangle are equal ​

Answer 1

$\huge \fbox \red{Answer}$

Theorem Statement:

Angle opposite to equal sides of an isosceles triangle are equal.

Proof:

Given, an Isosceles triangle ABC, where the length of side AB equals the length of side AC.

Therefore, AB = AC

Construction:

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

Solution⤵

Given, an Isosceles triangle ABC, where the length of side AB equals the length of side AC.

Therefore, AB = AC

Construction:

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.