It is known that in the triangle ABC, AC BC, the angle ACB is 90 degrees, D is the midpoint of AB, a

1.Proof: acb = 90°

ac⊥bcbf⊥ce

ace=∠cbg

aec=∠adc+∠dce=90°+∠dce，∠bgc=∠gfc+∠dce=90°+∠dce

aec=∠bgc

ac=bc△ace≌△cbg

AE = Proof: BF CH, AC BC

ach=∠cbf

ac=bcrt△ach≌rt△cbf

CH = BGAC = BC, the midpoint of AB at D.

cd⊥ab∠hcm=∠fbe

rt△chm≌rt△bfe

be=cm

1): Prove that the triformal AEC is equal to the triangular BGC

The reason is (AAS).

So: ae=cg

2）be=cm

Prove that the triangle BFE is equal to the triangle CHM

1.Proof: acb = 90°

ac⊥bcbf⊥ce

ace=∠cbg

aec=∠adc+∠dce=90°+∠dce,∠bgc=∠gfc+∠dce=90°+∠dce

aec=∠bgc

ac=bcace≌△cbg

AE = CG proof: BF CH, AC BC

ach=∠cbf

ac=bcrt△ach≌rt△cbf

CH = BGAC = BC, the midpoint of AB at D.

cd⊥abhcm=∠fbe

rt△chm≌rt△bfe

BE=cm,2,(1): Prove that the triformal AEC is fully equal to the triangle BGC on the grounds of (AAS).

So: ae=cg

2）be=cm

Prove that the triangle BFE is fully equal to the triangle chm,2,

1.Proof: acb = 90°

ac⊥bcbf⊥ce

ace=∠cbg

aec=∠adc+∠dce=90°+∠dce,∠bgc=∠gfc+∠dce=90°+∠dce

aec=∠bgc

ac=bc△ace≌△cbg

AE = Proof: BF CH, AC BC

ach=∠cbf

ac=bcrt△ach≌rt△cbf

CH = BGAC = BC, the midpoint of AB at D.

cd⊥ab∠hcm=∠fbe

rt△chm≌rt△bfe

be=cm

There are two situations for this question: The answer is as follows.

Sidehands. <>

1.Proof: acb = 90°

ac⊥bcbf⊥ce

ace=∠cbg

aec=∠adc+∠dce=90°+∠dce，∠bgc=∠gfc+∠dce=90°+∠dce

aec=∠bgc

ac=bc△ace≌△cbg

AE = CG proof: BF CH, AC BC

ach=∠cbf

ac=bcrt△ach≌rt△cbf

CH = BGAC = BC, the midpoint of AB at D.

cd⊥ab∠hcm=∠fbe

rt△chm≌rt△bfe

be=cm

1.Proof: acb = 90°

ac⊥bcbf⊥ce

ace=∠cbg

aec=∠adc+∠dce=90°+∠dce，∠bgc=∠gfc+∠dce=90°+∠dce

aec=∠bgc

ac=bc△ace≌△cbg

AE = CG proof: BF CH, AC BC

ach=∠cbf

ac=bcrt△ach≌rt△cbf

CH = BGAC = BC, the midpoint of AB at D.

cd⊥ab∠hcm=∠fbe

rt△chm≌rt△bfe

be=cm