In fact, this question is for an important property: the two bisectors of the two internal angles of a triangle intersect with the bisector of the internal angle of the third internal angle!

F is FM⊥AB in M, FN⊥AC in N and FP⊥BC in P respectively.

∴∠BMF=∠BPF=90

BF bisection ∠∴∠mbf=∠pbf. DBC

At △BMF and △BPF, ∠MBF=∠PBF, ∠BMF=∠BPF, BF=BF.

∴△BMF≌△BPF

∴MF=PF

Similarly, it can be proved that △ PCF △ NCF.

∴PF=NF

∴MF=NF

In Rt△AMF and Rt△ANF, ∠ AMF = ∠ ANF = 90, MF=NF, AF=AF.

∴△AMF≌△ANF

∴∠DAF=∠EAF, that is, AF is divided into equal parts ∠DAE.

∵AF⊥DE

∴△ADE, AF is not only the bisector of ∠DAE, but also its height to the opposite side DE, and ∴△ ade is an isosceles triangle (of course, this conclusion can also be obtained by proving congruence).