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
14
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