Question

i)Prove that the angle opposite to equal side of a triangle are equal​

Answer 1

Answer:

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

Answer 2

Step-by-step explanation:

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.