In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. If E and F be the midpoints of AC and AB respectively, prove that BE = CF.

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$G i v e n : △ A B C is an isosceles triangle in which A B = A C . E and F are midpoints of A C and A B, respectively . To prove : B E = C F Proof : E and F are midpoints of A C and A B, respectively . = > A F = F B, A E = E C A B = A C = > \frac{1}{2} A B = \frac{1}{2} A C = > F B = E C ∠ A B C = ∠ A C B (angle opposite to equal sides are equal) = > ∠ F B C = ∠ E C B Consider △ B C F and △ C B E : B C = B C (common) ∠ F B C = ∠ E C B (proved above) F B = E C (proved above) B y S A S congruence property : △ B C F ≅ △ C B E B E = C F (corresponding parts of the congruent triangles)$

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In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. If E and F be the midpoints of AC and AB respectively, prove that BE = CF. Figure

In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. If E and F be the midpoints of AC and AB respectively, prove that BE = CF.

Figure