Mathematics in the first year of junior high school proves that the steps are clear and urgently nee

Proof: ad bisects bac, bad= cad. (properties of the angular bisector) bad= cad (verified), b= eac. (known).

Bad+ B= CAD+ EAC (Equation Properties) ADE= BAD+ B (One outer angle of a triangle equals the sum of its two non-adjacent interior angles) and DAE= CAD+ EAC.

ade=∠dae。

Proof: Because AD is the angular bisector of ABC.

So bad= dac

And because eac= b

So ade= b+bad

eac+∠dac

dae

ADC is the outer angle of the triangle ADB, so the angle b = angle cae, the angle bad=angle dac, so the angle ade = angle b + angle dab= angle cac = angle cae + angle cad = angle ead, it is proven.

Proof: because, angle b + angle bad=angle ade; Angular EAC + Angular DAC = Angular DAE;

Angular bad=angular dac; Angle EAC = Angle B;

So, angular ade = angular dae; Hope.

The angle ade is equal to the angle b plus the angle bad

Angular bads are equal to angular dac

Angle B is equal to angle ECA

So the angle DAE is equal to the angle DAC plus CAE is equal to the angle B plus the angle BCD is equal to the angle ADE

Link BC, AD

ab = 8 cm, m is the midpoint of ab.

am=bm=4cm

The circumferential angles of the same arc are equal.

mcb=∠nad，∠cbm=∠adm

and BMC= DMA (equal to the vertex angle).

bcm~△dam

bm:cm=md:am

cm×dm=am×bm=16

cm:md=1:4

dm=4cm

cm×dm=4cm2=16

The solution yields cm=2cm, dm=8cm

cd=cm+dm=2+8=10cm

Solution: Connect BC and AD

It can be known that the angle bcd = the angle bad

Because angular cmb = angular amd

So there is a triangle CMB similar to the triangle AMD.

So cm ma = bm md

Substitution is x 4 = 4 4x

The solution is x=2, so cd=10

Solution: a b

1 = 3 (mark the corner of the letter d as 3) c d

The first question proves to be correct.

In the second question, mark the intersection of AG and CE as M, and mark the intersection of AG and BC as N

bag=∠bce

ad∥bc∠dag=∠cnm

bad=90°

bag+∠dag=∠bce+∠cnm=90°∴∠cmn=180°-90°=90°

ag⊥ce

The first sub-question de bf

The second sub-question de bf

Procedure: The sum of the internal angles of the quadrilateral is equal to 360, a= c=90, so adc+ abc=180

again abc + cbg = 180

So cbg = adc, and de, bf divide adc and cbg equally, so edc = cbf

Dec= beh. So bhe= c=90°, so the second sub-question of de bf is known from the previous question cbg= adc, which can get 1+ 2=90°, and then from c=90° to get 3+ 4=90°

1+ 2+ 3+ 4=180° is obtained, thus obtaining de bf