In Figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ΔABC ≅ΔDEF.
Given: BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC.
In ΔABC and ΔDEF
BA = DE (given)
BF = EC(given)
∠A = ∠D (both 90°)
BC = BF + FC
EF = EC + FC = BF + FC (∵ EC = BF)
⇒ EF = BC
Hence, ΔABC ≅ΔDEF (by RHS)