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).