As shown in Fig. 3, in the triangle ABC, ad bisects ∠ BAC, passing through B is vertical ad, perpendicular to foot is point E, crossing e is EF / / AC, and ab is intersected with F, AF and BF

∠AEB＝90º,∠FEA＝∠EAC＝∠EAB,∴ AF＝FE.
∠ABE＝90°-∠BAE＝90°-∠AEF＝∠FEB.∴FE＝FB,AF＝BF

As shown in the figure, △ ABC is an isosceles triangle, ∠ ACB = 90 °, extend BC to D, connect ad, pass point B as be ⊥ ad in E, and cross AC to F. in this figure, which two triangles can be regarded as one of the triangles rotated along a certain point? Try to explain why

∵ △ ABC is an isosceles triangle, ∵ ACB = 90 °,
∴AC=BC，∠BCF=∠ACD=90°，
And ∵ be ⊥ ad in E,
∴∠CBF=∠CAD，
∴△ACD≌△BCF，
Therefore, △ ACD is obtained by rotating △ BCF 90 ° clockwise around point C

As shown in the figure, in the triangle ABC, ∠ BAC = 90 ° Ao is perpendicular to BC to D, be bisects ∠ ABC and intersects ad with F. it is proved that the triangle AEF is an isosceles triangle

∠AFE=∠BFD=180-∠DBF-∠FDB
∠AEF=180-∠ABF-∠BAE
∠ DBF = ∠ ABF (angular bisector)
∠ FDB = ∠ BAE (right angle)
∠AFE=∠AEF
The triangle AEF is an isosceles triangle

As shown in the figure, in △ ABC, ad bisects ∠ BAC, de ∥ AC, EF ⊥ ad intersecting BC extension line at F. it is proved that ∠ fac = ∠ B

Proof: ∵ ad bisection ∠ BAC,
∴∠BAD=∠CAD，
∵DE∥AC，
∴∠EDA=∠CAD，
∴∠EDA=∠EAD，
∴AE=ED，
And ∵ EF ⊥ ad,
/ / EF is the vertical bisector of AD,
∴AF=DF，
∴∠FAD=∠FDA，
And ? fad = ∵ CAD + ∠ fac,
∠FDA=∠B+∠BAD，
∴∠FAC=∠B．

As shown in the figure, ad is the angular bisector of △ ABC, the extended line of be ⊥ ad crossing ad is at e, EF ∥ AC is crossing AB to F. it is proved that AF = FB

Proof: ∵ ad bisection ∠ BAC,
∴∠BAD=∠CAD，
∵EF∥AC，
∴∠FEA=∠CAD，
∴∠FAE=∠FEA，
∴FA=FE，
∵BE⊥AD，
∴∠FEA+∠FEB=90°，∠FBE+∠FAE=90°，
∴∠EBF=∠BEF，
∴EF=FB，
∴AF=FB．

As shown in the figure, in △ ABC, ad bisects ∠ BAC, de ∥ AC, EF ⊥ ad intersecting BC extension line at F. it is proved that ∠ fac = ∠ B

As shown in the figure, in △ ABC, ad bisects ∠ BAC, and the vertical bisector EF of ad intersects with the extension line of BC at point F and connects AF. it is proved that ∠ CAF = ∠ B

It is proved that: ∵ EF vertical bisection ad,  AF = DF,  ADF = ∠ DAF,
∵∠ADF=∠B+∠BAD，
∠DAF=∠CAF+∠CAD，
And ∵ ad bisection ∵ BAC,
∴∠BAD=∠CAD，
∴∠CAF=∠B．

As shown in the figure, in △ ABC, ad and BD bisect ∠ BAC and ∠ ABC, extend the circumscribed circle of △ ABC to e and connect be

It is proved that: ∠ EBC = ∠ EAC (equal to the circumference angle of the same arc). (2 points) ? ad, BD bisect  BAC and ∠ ABC, ? BAE = ∠ EAC, ∠ DBC = ∠ abd, (1 point) ? EBC = ∠ BAE, (1 point) ? EBC + ∠ DBC = ? EBD (as shown in the figure) ∠ BAE + ∠ a

As shown in the figure, in triangle ABC, angle c = 2, angle B, ad is the angular bisector of triangle ABC, and point E is on the vertical bisector of DB As shown in the figure in the triangle ABC, angle c = 2, angle B, ad is the angular bisector of triangle ABC, and point E is on the vertical bisector of DB. What is the size relationship between AB and AC + CD? Please explain the reasons

It is very simple that the vertical bisector of BD intersects AB in M, BD in N. because Mn bisects BD vertically, MB = MD ∠ B = ∠ MDB (three lines in one) ∠ amd = ∠ B + ∠ MDB because the angle c = 2 angle B, so ∠ C = ∠ AMD in △ AMD and △ ACD ∠ C = ∠ AMD ∠ CAD (AD is the angular bisector of triangle ABC) ad = ad △ amd

As shown in the figure, it is known that ad is the bisector of the outer angle ∠ EAC of △ ABC, and ad intersects the circumscribed circle of triangle ABC at D. It is proved that DB = DC If you want the auxiliary line, it is better to attach the figure,

Let p be a point on the extension line of Da, angle ead = angle PAB = angle DCB, angle ead = angle CAD = angle CBD
I did it in happy practice

As shown in Fig. 3, in the triangle ABC, ad bisects ∠ BAC, passing through B is vertical ad, perpendicular to foot is point E, crossing e is EF / / AC, and ab is intersected with F, AF and BF