Question

prove that angles opposite to equal sides of an isosceles triangle are equal​

Answer 1

Isosceles Triangle Theorems and Proofs

Theorem 1: Angles opposite to the equal sides of an isosceles triangle are also equal.

Proof: Consider an isosceles triangle ABC where AC = BC.

We need to prove that the angles opposite to the sides AC and BC are equal, that is, ∠CAB = ∠CBA.

Isosceles Triangle

We first draw a bisector of ∠ACB and name it as CD.

Now in ∆ACD and ∆BCD we have,

AC = BC                                                                (Given)

∠ACD = ∠BCD                                                    (By construction)

CD = CD                                                               (Common to both)

Thus,  ∆ACD ≅∆BCD                                        (By SAS congruence criterion)

So, ∠CAB = ∠CBA                                              (By CPCT)

Hence proved.

Theorem 2: Sides opposite to the equal angles of a triangle are equal.

Proof: In a triangle ABC, base angles are equal and we need to prove that AC = BC or ∆ABC is an isosceles triangle.

Isosceles Triangle Theorem 2

Construct a bisector CD which meets the side AB at right angles.

Now in ∆ACD and ∆BCD we have,

∠ACD = ∠BCD                                                    (By construction)

CD = CD                                                               (Common side)

∠ADC = ∠BDC = 90°                                          (By construction)

Thus, ∆ACD ≅ ∆BCD                                         (By ASA congruence criterion)

So, AC = BC                                                         (By CPCT)

Or ∆ABC is isosceles.