AD is the bisector of the triangle ABC, DE AB is in E, DF AC is in F to verify AD bisector EF

Ask EF to be handed over to P

Because an bisects bac, de ab, df ac has de=df

aed≌△afd

So: ae=af

Again: ae af = ep pf

So: ep=pf

So AD divides EF equally

Because AD is an angular bisector, the angle EAD and the angular DAF are the same.

And because de ab is in e, and df ac is in f, the angle dea = angle dfa = 90 degrees.

And because ad=ad.So triangle AED and triangle AFD congruence (corner edge theorem, also angle bisector theorem).

So ae=af lets ef and ad intersect at point h.

And because the angle EAD and the angle DAF are the same, ae=af, ah=ah, the triangle AEH and the triangle AFH are congruent (corner edge theorem).

So EH = EF angle AHE = angle AHF and because angle AHE + angle AHF = 180 degrees.

So the angle AHE = angle AHf = 90 degrees and EH = FH So ah bisects ef i.e. ad bisects ef

Because: ad is the bisector of the triangle ABC.

de⊥ab,df⊥ac

So: de=df

Because: ad=ad, de=df, are all right triangles.

So: ADE and ADF congruence.

ae = af, therefore: angle aeg = angle afg

Provable: AEG and AFG congruence.

Angular age = angular agf

Angular age + angular agf = 180°

So: angular age = angular agf = 90°

ad⊥ef

(Let AD intersect EF at the point G).

AD bisects the BAC

bad=∠cad

aed= afd=90°, ad=ad

aed≌△afd（aas）

ed=fd，∠ade=∠adf

dg=dg

edg≌△fdg(sas)

egd=∠fgd=90°，eg=fg

AD bisects EF vertically

AD is the angular bisector.

bad=∠cad

de⊥ab,df⊥ac

dea=∠dfa=90° ade=∠adf∵ad=ad

ade≌△adf(aas)

ed=df ade= adf od=do(temporarily called o) ode odf(sas).

oe=of doe=∠dof

AD is the mid-perpendicular line of EF.

Because AD bisects the angle BAC, so DE=DF, and because DE and DF are high, so the 3-angular AED is all equal to the 3-angular AFD, so the angle EDA=angular FDA, so the 3-angular EDP is all equal to the 3-angular DFP, so EP=FP, the angle EPD=angular FPD=90, so AD bisects EF perpendicularly

AD bisects BAF, de AB, DF AC, EAD= FAD, DE=DF, and the point D is on the perpendicular bisector of EF. dea dfa, ae=af, point a is on the perpendicular bisector of ef. AD bisects EF vertically

The congruence of triangles is HL's theorem.

Pick mine, I'm the first!

AD bisects the BAC

bad=∠cad

aed= afd=90°, ad=ad

aed≌△afd（aas）

ed=fd，∠ade=∠adf

dg=dg

edg≌△fdg(sas)

egd=∠fgd=90°，eg=fg

AD bisects EF vertically

Corner bisector bad= cad

de⊥ab,df⊥ac

dea=∠dfa=90°

ad=ad△ade≌△adf

ae=af i.e. aef is an isosceles triangle.

AD is the angular bisector.

AD bisects EF vertically

For the sake of convenience, I have used 1, 2, 3, 4, and 5 to represent the corners in the diagram.

Which corner it refers to, just look at the picture.

Proof: Because AD bisects the angle BAC, the angle BAD=ANGLE DAC, because AB is parallel DE, so the angle BAD= ANGLE EDA, because AD is parallel EF, so the angle DAC= angle FEC, the angle ADE= angle DEC, so the angle DEF= the angle FEC, so EF bisects the angle DEC

Let ad and ef intersect at point O

Because EF bisects AD perpendicularly, so AE is equal to DE, AF is equal to DF, AOE is equal to DO, AOE and other friends are trembling at 90 degrees Because AD is the angle bisector of the angle BAC in the triangle ABC So BAD is equal to CAD Because AO is equal to AO so the triangle AOE is congruent with AOF, so AE is equal to AF So AE, AF, DE, and DF are equal So the four sides do the sensitive shape AEDF is a diamond.

Let CE and AD intersect at H, AE=AC, the triangle AEC is an isosceles triangle, AD is based on the three-in-one property of the isosceles triangle, AH is the high and middle line of the triangle AEC, that is, the perpendicular bisector of EC, ed=cd, EF BC, that is, CE bisect

Let EF and AD be handed over to O; AD is the angular bisector of the triangle ABC=> EAO= FAO, EF bisector AD=> AOE= AOf=90° RT AOE RTaof(ASA) =>AE=AF

rt△aof≌rtdof(sas) =>af=df；rt△aoe≌rtdoe(sas) =>ae=de

The quadrilateral aedf is a diamond.

EF AD is intersected at o, AD bisects bac, aoe aof, AD bisects EF, and EF bisects AD perpendicularly, so the quadrilateral aedf is a diamond and the quadrilateral diagonal bisects are diamonds

Hongtao (Hong: vigorous, prosperous).

From Rong, Chu Dan.

AD bisects the BAC

bad=∠cad

ae∥df∠bad=∠adf

cad=∠adf

af=dfab∥df,ac∥de

The quadrilateral aedf is a parallelogram.

AF=DF quadrilateral, AEDF is a diamond.

ad⊥ef