In Rt ABC, BAC 90 amp 176, BE bisected ABC, AD BC and BE in F, FC BC to verify AE CG

Proof: Passing the point E is made to eh perpendicular to H

So the angle BHE = angle che = 90 degrees.

Because the angle bac = 90 degrees.

So the angle bac = angle bhe = 90 degrees.

Because be bisected angle ABC

So the angle abe = the angle cbe

Because be=be

So the triangle abe and the triangle hbe are congruent (corner corners) so ae=eh

Because AD is perpendicular to BC over D

So the angle ADC = 90 degrees.

Because the angle ADC + angle C + angle DAC = 180 degrees.

So angle dac + angle c = angle bad+ angle diac = angle bac = 90 degrees.

So corner bad=angle c

Because the horn AFE = horn ABE + the horn bad

Angular AEF = Angular C + Angular Cbe

So the angle afe = the angle aef

So ae=af

So af=eh

Because FG is parallel to BC

So the angle AFG = angle ADC = 90 degrees.

Angle agf = angle c

So the angle AFG = angle EHC = 90 degrees.

So the right triangle AFG and the right triangle EHC congruence (AAS) so AG=EC

Because ag=ae+eg

ec=eg+cg

So ae=cg

Proof: Passing the point G as GH BE to hand over BC to H

ad⊥bc∠abc+∠bad＝90

bac＝90

abc+∠c＝90

bad＝∠c

Be bisected by ABC

abe＝∠cbe

aef＝∠c+∠cbe，∠afe＝∠bad+∠abe∠aef＝∠afe

ae＝afgh∥be

chg＝∠cbe

chg＝∠abe

fg∥bc，gh∥be

Parallelogram BFGH

bf＝gh△abe≌△chg （aas）

af＝cgae＝cg

The math tutoring team answered your questions, please understand in time.

Zhenghui infiltrated the slow, abc=45, cd ab, pre-mold dcb=45, db=dc

a+ abe=90, abe+ dfb=90 a= dfb, shout code dfb dac

bf=ac

It's done!!

bf/eb=da/ae

CE eb=ca ab (bisector theorem of internal angles).

On the other hand. dab=∠dac

cba=∠dca

Hence EAB DAC

Therefore, ca ab=da ae

Therefore, BF EB=CE EB, BF=CE

If you have any questions, please feel free to ask!!

Solution: Because in the RT triangle ABC, be bisects the angle abc and ed perpendicular to ab at the point d, so de=ec (the distance from any point on the angle bisector to both sides of the angle is equal), so ae+de=ae+ec=ac=3.

As eg perpendicular to ac in g, fh weeping straight in ab in h, fh=ed=eg, triangle cge and triangle mengcha bhf congruence, so ce=bf