The isosceles triangle is proved below.
The complete information wasn't found. The proofing of the triangle will be illustrated. It should be noted that the 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.
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)
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