As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad

As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad

This is the second point of the sum theorem
1. The outer angle of a triangle is greater than any inner angle which is not adjacent to it
2. The outer angle of a triangle is equal to the sum of two inner angles which are not adjacent to it
3. The sum of the outer angles of a triangle is 360 degrees
You see. Angle B + angle bad + angle ADB = 180 degrees
Angle ADB + angle ADF = 180 degrees
So the angle ADF = angle B + angle bad

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．

Ad is the bisector of the angle BAC in the triangle ABC, passing through the midpoint e of AD as the extension line of EF ⊥ ad crossing BC at F, connecting AF, and proving that angle B = angle caf

As shown in the figure: Fe is the vertical bisector of AD, so the triangle ADF is isosceles triangle; angle FDA = fad; and because angle FDA = DCA + DAC; fad = Fab + bad; because ad is the bisector of angle BAC, angle DAC = DAB; angle CAF = Fab + BAC; angle FBA = DCA + BAC

As shown in the figure, in the triangle ABC, ad bisects the angle BAC, intersects BC with D, points E and F are respectively on BD and ad, and EF is parallel to AB, ed = CD Sao Nian, the picture is in the webpage --

It is proved that eg ∥ AC is made and the extension line of ad is crossed at point G so that eg = EF
∵EG∥AC
∴∠GED＝∠C
In △ EGD and △ CAD
∠BED＝∠C
ED＝CD
∠EDG＝∠CDA
∴△EGD≌△CAD（ASA）
∴AC＝EG
∵EG＝EF
∴AC＝EF
(I don't know why. In the end, there are still a few conditions that are useless, but it should be right to do so)

As shown in Fig. 14-19, in the triangle ABC, AB > AC, ad bisect ∠ BAC, EF ⊥ ad in G, respectively intersect AB, AC in E, F, and intersect the extension line of BC at M. if the difference between ∠ ACB and ∠ ABC is 30 °, find ∠ M

According to the meaning of the title, ∠ AEF = ∠ AFE = ∠ MFC
∠AEF=∠B+∠M
∠AFE=∠NFC=∠ACB-∠M
Therefore, there is ∠ B + ∠ M = ∠ ACB - ∠ M
2∠M=∠ACB-∠B
∠M=½×30°=15°

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

It is proved that: ∵ ad ⊥ BC,  BDA = 90 °, ∵∵ BAC = 90 °,  ABC + ∠ C = 90 °, ABC + ∠ bad = 90 °, ∵ bad = ∠ C, ? an bisection ? DAC, ? can = ∠ Dan, ? ban = ∠ bad + ∠ Dan, ? be ⊥ an

In △ ABC, the angle BAC = 90 degrees, ad ⊥ BC in D, CE bisect angle ACB in G, intersection AB in E, EF ⊥ BC in F. it is proved that the quadrilateral aefg is a diamond

It is proved that: (1) the following proof is: (1) the ? C + DAC = 90 ∵ 8 ∵ 8 ? 8 ∵ 8 ? 8 ? be ∵ be ∵ be ∵ be ? be ? be ? be ? be ∵ be ? be ? be ? be ? a ? a BC ν ad / / EF ∵ Ag = EF ᙽ ag

In △ ABC, ∠ BAC = 90 ° ad ⊥ BC, D, CE ⊥ ABC, ad at point G, AB at point E, EF ⊥ BC with vertical foot f to prove that aefg is a rhombus person

∵ CE bisection angle ACB. ∵ BAC = 90 ° ad ⊥ BC
∴AE=EF ∠AEC=∠CEF
∵.∠BAC=90°AD⊥BC
∴AD∥EF
∴∠AGE=∠CEF
And ∵ AEC = ∠ CEF
∴∠AGE=∠CEF
∴AG=AE
∵.AE=EF
∴AG=EF
And ∵ ad ∵ EF, namely Ag ≓ EF
The quadrilateral aefg is a parallelogram
∵.AE=EF
The quadrilateral aefg is a diamond

It is known that in △ ABC, ∠ BAC = 90 °, ad ⊥ BC at point D, be bisection ∠ ABC, intersection ad at point m, an bisection ∠ DAC, intersection BC at point n Verification: the quadrilateral amne is rhombic

Proof: ∵ ad ⊥ BC,
∴∠BDA=90°，
∵∠BAC=90°，
∴∠ABC+∠C=90°，∠ABC+∠BAD=90°，
∴∠BAD=∠C，
∵ an bisection ∠ DAC,
∴∠CAN=∠DAN，
∵∠BAN=∠BAD+∠DAN，∠BNA=∠C+∠CAN，
∴∠BAN=∠BNA，
∵ be bisection ∵ ABC,
∴BE⊥AN，OA=ON，
Similarly, OM = OE,
The quadrilateral amne is a parallelogram,
The parallelogram amne is a diamond

In △ ABC, ab = AC, ad bisects the outer angle ∠ CAE of △ ABC

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