As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, de ⊥ AB is on e, f is on AC, BD = DF. The following results are obtained: (1) CF = EB; (2) CBA + ∠ AFD = 180 °

As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, de ⊥ AB is on e, f is on AC, BD = DF. The following results are obtained: (1) CF = EB; (2) CBA + ∠ AFD = 180 °

(1) In RT △ DCF and RT △ DEB, BD = dfdc = De, at △ DCF ≌ RT ≌ DEB (HL), CF = EB. (2) ≔ RT ≌ DCF ≌ RT ≌ DEB, (2) ≂ RT ≌ DFC = ∠ B. ≂ DFC + ∠ AFD = 180 degree, ≌ cab

As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, de ⊥ AB is on e, f is on AC, BD = DF. The following results are obtained: (1) CF = EB; (2) CBA + ∠ AFD = 180 °

(1) In RT △ DCF and RT △ DEB, BD = dfdc = De, at △ DCF ≌ RT ≌ DEB (HL), CF = EB. (2) ≔ RT ≌ DCF ≌ RT ≌ DEB, (2) ≂ RT ≌ DFC = ∠ B. ≂ DFC + ∠ AFD = 180 ° and ≌ cab + ∠ AFD = 180 °

As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, de ⊥ AB is on e, f is on AC, BD = DF. The following results are obtained: (1) CF = EB; (2) CBA + ∠ AFD = 180 °

(1) In RT △ DCF and RT △ DEB, BD = dfdc = De, at △ DCF ≌ RT ≌ DEB (HL), CF = EB. (2) ≔ RT ≌ DCF ≌ RT ≌ DEB, (2) ≂ RT ≌ DFC = ∠ B. ≂ DFC + ∠ AFD = 180 ° and ≌ cab + ∠ AFD = 180 °

In the triangle ABC, D is the midpoint of BC, DF ⊥ AC is in F, de ⊥ AB is in E, and EB = FC

DF ⊥ AC, de ⊥ AB, so ∠ DFC = ∠ DEB = 90
EB=FC
D is the midpoint of BC, BD = CD
So △ DFC ≌ Δ DEB.DF=DE
That is to say, the distance between AB and AC on both sides of D to ∠ BAC is equal
So D is on the bisector of ∠ BAC, ad bisectors ∠ BAC