ABC is an isosceles triangle in which the altitudes $B E$ and $C F$ are drawn to the equal sides $A C$ and $A B$ respectively. Then

B E = C F

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B E = A B

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A B = B C

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A C = B C

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Solution

The correct option is C $B E = C F$
In $△ A B E$ and $△ A C F$ ,

we have $∠ B E A = ∠ C F A$ (Each $90^{0}$ )

$∠ A = ∠ A$ (Common angle)

$A B = A C$ (Given)

$∴ △ A B E ≅ △ A C F$ (By SAS congruence criteria)

$∴ B F = C F$ [C.S.C.T]

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ABC is an isosceles triangle in which the altitudes BE and CF are drawn to the equal sides AC and AB respectively. Then

The correct option is C BE=CFIn △ABE and △ACF, we have ∠BEA=∠CFA (Each 900)∠A=∠A (Common angle) AB=AC (Given)∴△ABE≅△ACF (By SAS congruence criteria)∴BF=CF [C.S.C.T]

ABC is an isosceles triangle in which the altitudes $B E$ and $C F$ are drawn to the equal sides $A C$ and $A B$ respectively. Then

The correct option is C $B E = C F$
In $△ A B E$ and $△ A C F$ ,

we have $∠ B E A = ∠ C F A$ (Each $90^{0}$ )

$∠ A = ∠ A$ (Common angle)

$A B = A C$ (Given)

$∴ △ A B E ≅ △ A C F$ (By SAS congruence criteria)

$∴ B F = C F$ [C.S.C.T]