If ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively, then which of the following options is correct?

BE ≠ CF

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AF = CF

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BE = AE

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BE = CF

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Solution

The correct option is D BE = CF
$I n Δ A E B a n d Δ A F C ∠ A E B = ∠ A F C (e a c h 90^{\circ}) ∠ A = ∠ A (c o m m o n a n g l e) A B = A C (g i v e n) ∴ Δ A E B ≅ Δ A F C (b y A A S c o n g r u e n c e r u l e) \Rightarrow B E = C F (b y C . P . C . T)$

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If ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively, then which of the following options is correct?

The correct option is D BE = CFIn ΔAEB and Δ AFC∠ AEB=∠ AFC (each 90∘)∠ A=∠ A (common angle)AB=AC (given)∴ΔAEB≅ΔAFC(by AAS congruence rule)⇒BE=CF (by C.P.C.T)

The correct option is D BE = CF
$I n Δ A E B a n d Δ A F C ∠ A E B = ∠ A F C (e a c h 90^{\circ}) ∠ A = ∠ A (c o m m o n a n g l e) A B = A C (g i v e n) ∴ Δ A E B ≅ Δ A F C (b y A A S c o n g r u e n c e r u l e) \Rightarrow B E = C F (b y C . P . C . T)$