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Question
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.
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Solution
△ABC is an isosceles triangle.
∴ AB = AC
∠ACB = ∠ABC ...[Angles opposite to equal sides of a △ are equal]
∠BCE = ∠CBF
Now, in △BEC and △CFB
∠BCE = ∠CBF ...[Proved above]
∠BEC = ∠CFB ...[Each 90°]
BC = CB ...[Common]
∴ △BEC ≅ △CFB ...[By AAS congruence]
So, BE = CF ...[By Corresponding parts of congruent triangles]
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