As shown in the figure, in △ ABC, the angle BAC = 90 degrees, ad ⊥ BC bisects ∠ ABC at point D and BF, then △ AEF is an isosceles triangle? And the proof is given

∵AD⊥BC
∴∠ADB=90°
∵∠BFA=∠BAC-∠ABE=90°-∠ABE
∵ AEF = ∠ bed (equal to vertex angle)
And ? bed = ∠ ADB - ∠ CBF = 90 ° - ∠ CBF
∴∠AEF=90°-∠CBF
∵ BF bisection ∵ ABC
∴∠CBF=∠ABE
∴∠BFA=∠AEF
The AEF is an isosceles triangle

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

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, ∠ BAC = 90 ° ad ⊥ BC BF bisection is called ABC to prove that △ AEF is an isosceles triangle

∵∠BAF=∠BAC=90°
∴∠BFA+∠ABF=90°
∵AD⊥BC
∴∠ADB=90°
∴∠EBD+∠BED=90°
And ∵ bed = ∵ AEF
∴∠AEF+∠BED=90°
∵ BF bisection ∵ bac
∴∠ABF=∠DBF
∴∠AEF=∠AFE
∴AE=AF
That is, △ AEF is an isosceles triangle

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

As shown in the figure, ∠ BAC = 90 ° in △ ABC, ad ⊥ BC in D, BF bisection ∠ ABC crossing ad in E. (1) ∠ AEF and ∠ AFE are equal? Please explain the reasons（ (2) Compare the size relationship between ∠ C and ∠ bed, and explain the reason

1. In the case of the evidence that ∵ the ∵ that ∵ the ∵ BAC = 90 ∵ C ? ABC = 90 ? ad ? BC ? BC ? ABC ? ABC ? ABC ∵ bad = 90 ? bad ? C ? the ? of ? BAC ? BAC = 90 ? C ? C AEF = AFE AEF = ∠ AFE AEF = ∠ afe2, ∠ bed

As shown in the figure, it is known that in △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. it is proved that BF = 2cf

Proof: connect AF
∵AB＝AC,∠BAC＝120
∴∠B＝∠C＝（180-∠BAC）/2＝30
∵ EF vertical bisection AC
∴AF＝CF
∴∠CAF＝∠C＝30
∴∠BAF＝∠BAC-∠CAF＝90
∴BF＝2AF
∴BF＝2CF
The math group answered your question,

As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf

Proof: AF connection, (1 point)
∵AB=AC，∠BAC=120°，
∴∠B=∠C=180°−120°
2 = 30 °, (1 point)
∵ the vertical bisector EF of AC intersects AC at point E and BC at point F,
ν CF = AF (the distance from the point on the vertical bisector of the line segment to the point at both ends of the line segment is equal),
﹤ fac = ∠ C = 30 ° (equilateral and equal angle), (2 points)
Ψ BAF = ∠ BAC - ∠ fac = 120 ° - 30 ° = 90 °, (1 point)
In RT △ ABF, ∠ B = 30 °,
ν BF = 2AF (in a right triangle, the right angle to which the 30 ° angle is opposite is equal to half of the hypotenuse), (1 point)
ν BF = 2cf (equivalent substitution)

As shown in the figure, △ ABC, ab = AC, ∠ BAC = 120 ° and the vertical bisector EF of AC intersects AC at point E and BC at point F. verification: BF = 2cf

It is proved that: connect AF, (1 minute) ∵ AB = AC,  BAC = 120 °,  B = ∵ C = 180 ° − 120 ° 2 = 30 °, (1 minute)

As shown in the figure, in △ ABC, the angle BAC = 90 degrees, ad ⊥ BC bisects ∠ ABC at point D and BF, then △ AEF is an isosceles triangle? And the proof is given