prove that the opposite angle of two equal sides of an isosceles triangle are there same
Answers
Step-by-step explanation:
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.
Answer:
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Step-by-step explanation:
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