Q. 14
In Fig. 10.136, < A. Δ EFC
B. Δ ECF
C. Δ CEF
D. Δ FEC
Answer :
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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