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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 |
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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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