Prove:

In △BDE and △CDF

BD=CD

∠BED =∠CFD = 90°

BE=CF

∴△BDE≌△CDF(HL)

∴DE=DF

∵DE⊥AB

DF⊥AC

Virtue? DF is AD to AB respectively? The distance of AC

∴AD is the bisector of △ ABC ∠ a.

Ahem, please post the question next time, it's very painful to find. ...