In the equilateral triangle ABC, ad and CE are the heights on BC and ab respectively, and AD and CE intersect at point F, what is the degree of ∠ AFE? If points D and e run on BC and ab respectively, in order to make the conclusion (1) hold, please guess what quantitative relationship BD and AE should satisfy and give proof

∠ AFE = 60 ° ∵ △ ABC is an equilateral triangle, and ∵ ad is the height of BC  ad is the angular bisector of  EAF  CAF = 30 ? CE is the height of ab ? CEA = 90 ? AFE = 180 ° - 90 ° - 30 ℃ = 60 ℃ (2) BD = AE ∵ BD = AE, ab = AC ? abd 

Given that the triangle ABC is an equilateral triangle, the points de are on the sides of BC and AC respectively, and AE = CD, ad and be intersect at point F, and calculate the degree of angle BFD

Delta ABC is an equilateral triangle
∴ AC=AB,∠BAC=∠C=60º
∵ DC=AE
∴ △ADC≌△BEA
∴ ∠CAD=∠ABE
∵ BFD = ∠ BAF + ∠ Abe and ∠ CAD = ∠ Abe
∴ ∠BFD=∠BAF+∠CAD=∠BAC=60º

As shown in the figure, in △ ABC, ad bisects ∠ BAC to BC to D, be ⊥ AC to e, and ad to F. it is proved that ∠ AFE = 1 2（∠ABC+∠C）．

∵ the sum of the inner angles of the triangle is 180 degrees,
∴∠BAC=180°-（∠ABC+∠C），
∵ ad bisection ∵ BAC to BC to d,
∴∠DCA=1
2∠BAC=90°-1
2（∠ABC+∠C），
∵ be ⊥ AC to E,
∴∠AFE=90°-∠FAE=90°-90°+1
2（∠ABC+∠C）=1
2（∠ABC+∠C）．

In the triangle ABC, D is BC point, e is ad point, angle DAC is equal to angle B, CD is equal to CE, ACE and bad are similar

The ace of triangle is similar to bad
reason:
Because CD = CE
Therefore, CDE = CED
Because ∠ CDE ＋∠ ADB = 180 degrees, ∠ CED ∠ AEC = 180 degrees
So, ADB = AEC
Because ∠ B = DAC, i.e., B = EAC
So △ ace ∽ bad
For reference! (jswyc)

It is known that in the triangle ABC, be is an angular bisector and ad is perpendicular to D. It is proved that ∠ bad = ∠ DAE + ∠ C

Prolonged ad crossed BC to F
⊿ abd ⊿ FBD
∵∠ADB=∠FDB=90°,∠ABD=∠FBD,BD=BD
∴⊿ABD≌⊿FBD
∴∠BAD=∠BFD
∵ BFD = ∠ C + ∠ FAE (the outer angle of a triangle is equal to the sum of two inner angles not adjacent to it)
∴∠BAD=∠DAE+∠C

As shown in the figure, in the triangle ABC, ∠ B is greater than ∠ C, ad is the high angle on BC side, and AE bisection ∠ BAC proves ∠ DAC = 1 / 2 (∠ B - ∠ C)

In △ abd, AE ⊥ BD ⊥ ade = 90 ⊙ DAE,  BAE + ∠ B = 90 ° - ∠ DAE,  ad is the bisector of the angle BAC ? bad = ∵ DAC

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

As shown in the figure, in △ ABC, ∠ BAC = 60 °, B = 45 ° and ad is an angular bisector of △ ABC, then ∠ DAC= Degree, ∠ ADB= Degree

∵ ad is an angular bisector of ∵ ABC,
∴∠DAC=30°，
∴∠ADB=∠DAC+∠C=30°+（180°-60°-45°）=105°．
Therefore, fill in 30; 105

As shown in Figure 3-13, ad and BF are the bisectors of the height and angle of the triangle ABC respectively. BF and ad intersect at e, and the angle AFE = angle AEF

If FG is perpendicular to BC and the foot perpendicular to g, there are: AD / / FG, so: ∠ bed = ∠ BFG because ∠ AEF = ∠ bed (opposite vertex angle) ∠ AFE = ∠ AEF, and because: be is an angle bisector, so: ∠ ABF = ∠ GBF BF = BF, so triangle ABF is equal to triangle GBF, so ∠ FGB = ∠ BAC with BAC = 90 degrees

As shown in the figure, in △ ABC, ∠ BAC = 90 °, ad ⊥ BC in D, BF bisection ∠ ABC intersects ad at point E and AC at point F. it is proved that ∠ AEF ⊥ AFE

The crossing point F is FH ⊥ BC, and it intersects with BC at h ? EDB  FHB  EBD  FBH  bed = 90 ﹣ EBD ﹤ BFH = 90 ⊥ FBH ᚉ bed ∵ BFH ∵ AFB ∵ bed ∵ AEF ∵ AFE, please accept it in time