As shown in the figure, ab ⊥ AC, ab = AC, ad ⊥ AE, ad = AE

As shown in the figure, ab ⊥ AC, ab = AC, ad ⊥ AE, ad = AE

[if you don't know the diagram, set ad between ∠ BAC]
prove:
∵AB⊥AC
∴∠BAD+∠DAC=90º
∵AD⊥AE
∴∠CAE+∠DAC=90º
∴∠BAD=∠CAE
And ∵ AB = AC, ad = AE
∴⊿ABD≌⊿ACE(SAS)