Question

please send me the answer fast with full explanation​

Answer 1

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Given

• ∠E = ∠F = 90°
• BF = CF [Equal altitudes of triangle]

To Prove

• ∠B = ∠C

Proof

In △BFC and △CEB

BE = CF [ Given ]

BC = BC [ Common ]

∠BFC = ∠BEC [ each equal to 90° ]

So, △BFC ≅ △CEB [ By RHS congruence rule ]

∠B = ∠C [ by CPCT ]

Hence Proved ∠B = ∠C

Answer 2

Answer:

Given

∠E = ∠F = 90°

BF = CF [Equal altitudes of triangle]

To Prove

∠B = ∠C

Proof

In △BFC and △CEB

BE = CF [ Given ]

BC = BC [ Common ]

∠BFC = ∠BEC [ each equal to 90° ]

So, △BFC ≅ △CEB [ By RHS congruence rule ]

∠B = ∠C [ by CPCT ]

Hence Proved ∠B = ∠C