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Let us know that the triangle ABC, D is a point on BC, e is a point on AC, and BD = CE, connect ad, BD, intersect with point F, and find the degree of angle AFE
60 degrees
In the equilateral triangle ABC, D and E are on the sides BC and AC respectively, DC is equal to AE, ad, be intersects with point F, and the proof angle BFD is equal to 60 degrees
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
It is known that the triangle ABC is an equilateral triangle, points D and E are on the sides of BC and AC respectively, and AE = CD, ad and be intersect at point F. (1) it is proved that the triangle Abe is an equilateral triangle It is known that the triangle ABC is an equilateral triangle, points D and E are on the sides of BC and AC respectively, and AE = CD. AD and be intersect at point F （1） Prove that triangle ABE is equal to triangle CAD (2) Find the degree of angle BFD
(1) Delta ABC is an equilateral triangle
So AB = AC, ∠ BAE = ∠ ACD
AE = CD
△ABE≌△CAD
（2）△ABE≌△CAD
So ∠ CAD = ∠ Abe
The ∠ BFD is the external angle of △ fab,
So ∠ BFD = ∠ Abe + ∠ BAF = ∠ CAD + ∠ BAF = ∠ bad = 60 °
Δ ABC is an equilateral triangle, the points de are on the edge of BCAC respectively, and AE = CD, ad and be intersect at point F. find the degree of ∠ BFD
∵ equilateral △ ABC
∴AB＝AC,∠BAC＝∠C＝60
∵AE＝CD
∴△ABE≌△CAD （SAS）
∴∠ABE＝∠CAD
∴∠BFD＝∠BAD+∠ABE＝∠BAD+∠CAD＝∠BAC＝60°
Welcome to adopt
d. E is the point on the edge BC, AC in the equilateral triangle ABC, connecting ad, be, intersecting F, and the angle BFD = 60 ° respectively
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It is proved that △ Abe is equal to △ CAD and that △ Abe is equal to △ CAD It is proved that △ Abe is equal to △ CAD and the degree of angle BFD
（1）AB=AC,
AE=CD,
∠BAE=∠ACD=60,
∴△ABE≌△CAD（SAS）.
（2）△ABE≌△CAD,
∠EAF=∠ABE,
∠AFE=∠FBA+∠BAF
∠AFE=∠FAB+∠EAF=∠BAE=60,
∠BFD=∠AFE
∠BFD=60°.
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）．
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）．
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∠AFE=∠BFD
∠BFD=∠ABF+∠BAD
1/2（∠ABC+∠C）=1/2（180-∠A)=90-1/2∠A=∠ABF+∠BAD
So ∠ AFE = 1 / 2 (∠ ABC + ∠ C)
In △ ABC, ∠ BAC = 90 °, C = 30 ° and the height ad intersects the bisector be of ∠ ABC at F. it is proved that △ AFE is an equilateral triangle
prove:
∵∠C=30°,AD⊥BC
∴∠CAD=60°,∠ABC=60°
∵ be bisection ∵ ABC
∴∠CBE=30°
∴∠AEF=∠C+∠CBE=60°
∴∠EAF=∠AEf=60°
△aef is an equilateral triangle
RELATED INFORMATIONS
- 1. 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）．
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