It is known that BC DE, angle B angle E, angle C angle D, F are the midpoints of the CD to verify th

Proof: Connect BF and EF. Connect BE.

bc=ed,cf=df,∠bcf=∠edf。

BCF EDF (Corner Edge).

bf=ef,∠cbf=∠def。

In BEF, BF=EF. ∴ebf=∠bef。and abc= aed.

abe=∠aeb。

ab=ae。

In ABF and AEF, AB=AE, BF=EF, ABF= ABE+ EBF= AEB+ BEF= AEF= AEF.

abf≌△aef

baf=∠eaf。

Proof: Connect BF and EF.

bc=ed,cf=df,∠bcf=∠edf。

BCF EDF (Corner Edge).

bf=ef,∠cbf=∠def。

Connect BE. In BEF, BF=EF. ∴ebf=∠bef。and abc= aed.

abe=∠aeb。

ab=ae。

In ABF and AEF, AB=AE, BF=EF, ABF= ABE+ EBF= AEB+ BEF= AEF= AEF.

abf≌△aef

baf=∠eaf。

First of all, it is used as an auxiliary line, connecting bf, linking, ab=de, and Jingchang a= d (corner edge theorem) so the triangle abf dec, we get the edge bf ce, and afb= dceAnd because the side BC=FE, and the chaotic edge BF=CE derived from the congruence of the triangle (the quadrangle with equal sides is a parallelogram), the quadrilateral.

Solution c= b (known).

EFC and DFB are opposite apex angles.

EFC = DFB (defined for apex angle).

The sum of the inner angles of the triangle is 180°

cef=180°-∠c-∠cfe

fdb=180°-∠b-∠bfd

cef = fdb (property of the equation).

CEF+ FEA = 180° (flat angle definition).

The same can be said. fdb+∠fda=180°

FEA = FDA (Properties of Equations).

In AEF and ADF.

1 = 2 (known).

FEA = FDA (verified).

af=af(common edge).

aef≌△adf（aas）

df = ef (congruent triangles correspond to equal sides).

Connect to the AC AD

It is known that ABC is fully equal to AED (SAS).

So ac=ad angle bac= angular dae

Because cf = df af = af

ACF is fully equal to ADF(SSS).

So angular caf = angular daf

Angular AFC = Angular AFD = 90°

So the angle BAF = BAC + CAF = DAE + DAF = Angle EAF (1) AF is perpendicular to CD (2).

Proof: The angle efc = angle DFB obtained by the equivalence of the top angles, and the angle AFC = angle AFB because the angle 1 = angle 2, and the angle b = angle c, and the common edge af, which is obtained by the angle edge theorem.

Triangle AFB Triangle AFC

So the angle fab = angle fac, af this common side, angle 1 = angle 2, which is obtained by the angle edge angle theorem.

Triangle AFE Triangle AFD

EF=df

Because b=142°

bfe=38°

It is known) so b+ bfe=142°+38°=180° definition of the angle of flat coarse objects).

So ab ef

The inner angles of the same side are complementary, and the two rocks are liquid in a straight line).

And because d=140°, efd=40°

Known) so d+ efd=140°+40°=180° flat angle definition).

So old rough to EF CD

The same side inner angle is complementary, and the two straight lines are parallel).

Because ab ef

EF cd so ab cd

Equivalent substitution).

Proof: b=142°

BFE = 38° (known).

b + bfe = 142 ° + 38 ° = 180 ° (circle only guess the equation property) ab ef (the same side of the inner angle is complementary, two straight lines are parallel).

d = 140°, efd = 40° (known) orange type.

d+ mountain acres efd = 140° + 40° = 180° (equilibrium properties) ef cd (complementary to the side of the inner angle, two straight lines are parallel).

ab cd (two lines parallel to the same line).

In the figure, points E and C are on BF, BE=FC, ABC= DEF=45°, A= D=90°.

If AC intersects de at m, and ab = root number 3, me = root number 2, rotate the line segment ce clockwise around point c so that point e rotates to g on ab, and find the length of gb.

1、∵be=fc

be+ec=ec+fc

i.e. bc = ef

abc=∠def=45°，∠a=∠d=90°∴△abc≌△def

ab=df2, abc= def=45°, a= d=90° abc and def are isosceles right triangles.

ab=ac=√3

acb=∠def=45°

i.e. mec= mce=45°

emc=90°，me=mc

EMC is an isosceles right triangle.

em=cm=√2

ec=2 (Pythagorean theorem em +cm = ec 2 + 2 = ec).

ec=gcgc=2

ab=acac=√3

gc = 2ag = 1 (Pythagorean theorem CG = AC + AG2 = 3 + AG).

gb=√3-1