In Fig. 10.136, AB ⊥ BE and FE ⊥ BE. If BC=DE and AB=EF, then Δ ABD is congruent to

In
ABD and
FEC,
AB= FE (Given)
∠B = ∠E (Each 90°)
BC = DE (Given)
Add CD both sides, we get
BD = EC
Therefore, by S.A.S. theorem,
ABD
FEC
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