Question

1.     To prove Angles opposite to equal sides of an isosceles triangle are​

Answer 1

Answer:

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.

Step-by-step explanation:

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Answer 2

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