Prove that angles opposite to equal sides of an isosceles triangle are equal.
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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.Nov
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let there be a triangle ABC, in which AB=AC.
Construct AP bisecting angle A meeting BC at P.
to prove- angles opposite to equal sides of an isosceles triangle are equal
proof-
In ∆ABP and ∆ACP
AP=AP(common)
AB=AC(given])
angle BAP=angle CAP (by construction])
Therefore, ∆ABP congurent ∆ACP by SAS
This implies, angle ABP=angleACP by CPCT
Hence proved
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