Given in figure,
BA⊥AC,DE⊥DF such that BA = DE and BF = EC
To show : ΔABC≅ΔDEF
Proof :
Since BF =EF
On adding CF both side, we get
BF +CF = EC +CF
⇒BC=EF...(i)
In ΔABC and ΔDEF∠A=∠D=90∘[∵BA⊥AC and CE⊥DF]
BC = EF [from Eq.(i)]
and AB = DE [given]
∴ΔABC≅ΔDEF [by RHS congruence rule]