Question

SHOW THAT THE ANGLES OPPOSITE TO EQUAL SIDES OF A TRIANGLE ARE EQUAL​

Answer 1

Step-by-step explanation:

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

This result can be proved in many ways. One of the proofs is given here.

Proof :

We are given an isosceles triangle ABC in which AB = AC. We need to prove that ∠B = ∠C.

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

In Δ BAD and Δ CAD,

AB = AC (Given)

∠BAD = ∠CAD (By construction)

AD = AD (Common)

So, Δ BAD ≅ Δ CAD (By SAS rule)

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

So, ∠B = ∠C

U can try this experiment too !!!

Construct a triangle ABC with BC of any length and ∠ B = ∠ C = 50°. Draw the bisector of ∠ A and let it intersect BC at D .Cut out the triangle from the sheet of paper and fold it along AD so that vertex C falls on vertex B.What can you say about sides AC and AB? Observe that AC covers AB completely So, AC = AB

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