1.

In Fig., AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then Δ ABD is congruent toA. Δ EFCB. Δ ECFC. Δ CEFD. Δ FEC

Answer»

In ΔABD and ΔFEC,

AB = FE (Given)

∠B = ∠E (Each 90°)

BC = DE (Given)

Add CD both sides, we get

BD = EC

Therefore, by S.A.S. theorem,

ΔABD ≅ ΔFEC



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