Question

In figure AB ⊥ BE and EF ⊥ BE. If BC = DE and AB = EF, then ΔABD is congruent to

Answer 1

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