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In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE，
∵∠APE=∠ABE+∠BAD，∠ABE+∠CBE=60°，
∴∠APE=∠ABC=60°．
Therefore, D

As shown in the figure, in the right triangle ABC, the angle BAC is equal to 90 degrees, AB is equal to AC, BD is the angular bisector of angle ABC, CE is perpendicular to BD, and intersects the extension line and point e of BD. it is proved that BD is equal to 2ce

The bisector of BAC is BD, CE ⊥ be.  be.  be.  be.  be.  be.  MBE = ≌≌≌ EM = ce

As shown in the figure, in △ ABC, ∠ a = 90 ° AB = AC, the bisector BD of ∠ ABC intersects AC at D, and the extension of CE ⊥ BD is at point e. it is proved that CE = 1 2BD．

Extend CE and Ba intersect at point F
∵∠EBF+∠F=90°，∠ACF+∠F=90°
∴∠EBF=∠ACF．
In △ abd and △ ACF
∠EBF＝∠ACF
AB＝AC
∠BAC＝∠CAF
∴△ABD≌△ACF（ASA）
∴BD=CF
In △ BCE and △ BFE
∠EBF＝∠CBE
BE＝BE
∠CEB＝∠FEB ，
∴△BCE≌△BFE（ASA）
∴CE=EF
∴CE＝1
2CF＝1
2BD．

If the bisector BD and CE intersect at point I in triangle ABC, can the degree of angle BIC be four times that of angle a?

In order to express conveniently, angle ABC=2x, angle ACB=2y. then angle A=180-2x-2y, x+y= (180- angle A) /2. then angle BIC=180-x-y=180-[(180- angle A) /2]=90+ (angle A/2). Make angle BIC=4* angle A, substitute into 90+ (angle A/2) =4* angle A, find out angle A= (180/7) degree, angle bic= (720/7) degree, and all conform to

As shown in the figure, it is known that △ ABC is an equilateral triangle, D is a point on the edge of BC, with AD as the edge, ∠ ade = 60 ° is made, and the bisector ce of the outer corner of △ ABC intersects at point E, Connect AE. Judge the shape of △ ade and prove your conclusion Be more specific

(1) Proof:
As shown in the figure, cut BH = BD on ab
⊿ ABC is an equilateral triangle
∴∠B=60,ZB＝AC,∠ACB＝60
And ∵ BH = BD
∴AH=DC
∵ CE bisects the external angle of ᙽ ACB, and  ACB = 60
∴∠ACE＝60
∴∠DCE＝∠ACB＋∠ACE＝120
∵∠B＝60,BH=BD
⊿ Bhd is an equilateral triangle
∴∠BHD=60
∴∠AHD＝120
∴∠AHD=∠DCE
∵∠ADC＝∠ADE＋∠EDC
And ∠ ADC = had + B
∴∠ADE＋∠EDC＝∠HAD＋∠B
And ∵ ade ∵ B = 60?
∴∠HAD＝∠EDC
In ⊿ ahd and ⊿ DCE
｛∠HAD＝∠EDC
｛∠AHD=∠DCE
｛AH=DC
∴⊿AHD≌⊿DCE（AAS）
∴AD＝DE
（2）
As shown in the figure, BH = BD is intercepted on the extension line of ab
⊿ ABC is an equilateral triangle
∴∠2＝∠1＝60,AB=BC,∠ABC=60
And ∵ BH = BD
⊿ ah = CD and ⊿ BDH are equilateral triangles
∴∠H＝60,∠BDH＝60
And ∵ CE bisects the external angle of ∵ ACB, and ∵ ACB = 60
∴∠3＝60
∴∠3＝∠H
∵∠ADH=∠ADE＋∠BDH－∠4＝120－∠4
And ∠ Dec = 180 - ∠ 3 - ∠ 4 = 120 - ∠ 4
∴∠ADH＝∠DEC
⊿ in ⊿ ahd and ⊿ DCE
｛∠3＝∠H
｛∠ADH＝∠DEC
｛AH=CD
∴⊿AHD≌⊿DCE（ASA）
∴AD=DE

As shown in the figure, AD.CE Is the angular bisector of the triangle ABC, AD.CE Intersect at point F. known As shown in the figure, AD.CE It's the angular bisector of △ ABC, AD.CE Intersect at the point F. it is known that the angle is B=60 ° and the confirmation is ae+cd=ac

If AG is intercepted on AC so that Ag = AE and FG is linked, Δ AGF ≌ Δ AEF
∠A+∠C=180-60=120º,∴(∠A+∠C)/2=60º
∴∠AFC=180-60=120º,∴∠EFD=120º
∴∠AFE=∠DFC=[360-(120+120)]/2=60º
∴∠AFG=∠AFE=60º,∴∠GFC=120-60=60º=∠DFC
∴ΔGFC≌ΔDFC,∴CG=CD
∴AE+CD=AG+CG=AC

As shown in the figure, △ ABC is an equilateral triangle, points D and E are on BC and AC respectively, and BD = CE, ad and be intersect at point F (1) It means △ abd ≌ △ BCE (2) Is △ AEF similar to △ Abe? Why not (3) Is BD? 2 = ad * DF? Please explain the reason

(1) Because of the equilateral triangle ABC, ab = BC, ∠ abd = ∠ BCE because BD = CE, ∠ abd = ∠ BCE, ab = BC, so △ abd ≌ △ BCE (2) because △ abd ≌ △ BCE, so ∠ bad = ∠ CBE because ∠ BAC = ∠ CBA = 60 ° so ∠ EAF = ∠ EBA because ∠ AEF ≌ △ bea (3) because ∠ DBF

D. E is the edge of the equilateral triangle ABC BC.AC It is proved that the angle AFE is 60 degrees I only prove that the triangle abd and the triangle BCE congruence (SAS), but not later,

It is proved that triangle ACD and triangle BAE are congruent
The method is the same as triangle abd and triangle BCE
So there is angle CAD = angle Abe
In triangle Abe, the sum of inner angles is 180 degrees, that is, angle Abe + angle AEB = 180 degrees - angle EAB (60 degrees) = 120 degrees
Then the angle CAD + angle AEB = 180 degrees - angle EAB (60 degrees) = 120 degrees
In triangle AEF, angle EAF (CAD) + angle AEF (AEB) = 120 degrees
So the angle AFE = 60 degrees

D E is the point on the edge BC and AC of the equilateral triangle ABC respectively, and BD = CE, connected be = CE, they intersect the point F, and calculate the degree of angle AFE

60 degrees

It is known that, as shown in the figure, in the equilateral triangle ABC, the point D E is on BCAC respectively, BD = CE, connecting ad, be, and intersecting with point F, and it is proved that ∠ AFE = 60 °

Certificate:
Because the equilateral triangle ABC, BD = CE, ∠ ABC = ∠ ACB,
Then △ abd is all equal to △ BCE;
∴∠BDA=∠BEC,∠FBD=∠BAD,
∵ sum of internal angles of triangle = 180 °,
∴∠BFD=∠ABD,
The △ BDF is similar to △ bec,
﹤ BFD = ∠ BCE = 60 ° = ∠ AFE (equal vertex angle)