A90, B90, CD AD BC, E are known to be the midpoints of AB

Last time, I proved a problem for you, and now it's here again.

This is very simple, if you don't know anything in the future, ask me directly.

Method: Proved with the Pythagorean theorem.

Analysis: Just prove that de2 +ce2 = cd2 don't look at the following words It's actually very simple.

Just to make you understand that there are more analysis parts Note: The 2 after the letter indicates the square.

Steps: 1. If you make DF BC and F then you have CD2=DF2+CF2

DF=AB=AE+EB=2AE (easy to see) CF=BC-AD

cd2=df2 +cf2 =（2ae）2 +（bc-ad）2 =4ae2 +ad2 +bc2 -2ab×bc ①

2. CD=AD+BC (given in the question) So CD2=(AD+BC) 2=AD2+BC2+2AD BC

From 4AE2 +AD2 +BC2 -2AB BC= AD2+BC2+2AD BC (i.e. equality after the two equations), AE2 = AD BC can be obtained

3. ADE is a right triangle so ed2= AD2+AE2

In the same way, EC2=BE2+BC2

ed2+ ec2 = ad2+ae2+ be2+bc2 (ae=be)

ad2 +bc2+2ae2

The above has already concluded that ae2 =ad bc, so it is enough to change the square of ae.

ed2+ ec2= ad2 +bc2 +2 ad×bc= cd2

It is concluded that de2 +ce2 =cd2 so dec is a right triangle.

Draw a line EF parallel to BC at E and F on DC.

Since e is the midpoint of ab, then f is the midpoint of dc.

Due to ad||bc||EF so EF=(AD+BC) 2 Known cd=ab+BC So DF=EF=FC so FD= Fed

Since ade= fed, then ade= fde, the same goes for bce= fce

Since adc + bcd = 108 so edc + ecd = 90 because the sum of the inner angles of the triangle is 180 so dec = 90 i.e. de ec

A lot of theorems, axioms and so on have been forgotten! If you think it's fun, practice your hands.

Use the vector method to establish a Cartesian coordinate system to solve.

Method 1: Make a flat group through the point E and cross the parallel lines of AB BC and F, because AB is parallel to CD to get AB, CD, EF three lines are parallel, so the angle BEF = 1 = 2, the angle CEF = 4 = 3, the triangle BEF and the triangle CEF are the isosceles triangle of filial piety, EF=BF=CF, that is, F is the midpoint of BC, that is, EF is the quadrilateral ABCD (because of coincidence or silver is AB...).

Because the empty auspicious is C, it is the midpoint of AB.

So ac=bc

In triangular ADCs and triangular BECs.

Because AC=BC (Proven Calendar) AD=BE, CD=CE (known), the triangle ADC congruence fights fiercely against the triangle BEC

So angle a = angle b (property of congruent triangle).

Method 1: Make a parallel line parallel to AB through the point of E to cross BC and F, because AB is parallel to CD to get AB, CD, EF three lines are parallel, so the angle BEF = 1 = 2, the angle CEF = 4 = 3, the triangle BEF and the triangle CEF are isosceles triangles, EF=BF=CF, that is, F is the midpoint of BC, that is, EF is the median line of the quadrilateral ABCD (because AB is parallel CD, then this quadrilateral is not a parallelogram or trapezoid), and 2EF=AB+ can be obtainedcd=bc

Method 2: Proof of:

Extension of the CE BA extension to F

E is the AD midpoint.

ef=ecab//cd

f= 4 and aef= dec

aef≌△dec

af=dc∠f=∠4，∠3=∠4

f=∠3bf=bc

i.e. BA+AF=BC

i.e. BC=AB+CD

Proof: Extension of CE to BA extension to F

E is the AD midpoint.

ef=ecab//cd

f= 4 and aef= dec

aef≌△dec

af=dc∠f=∠4，∠3=∠4

f=∠3bf=bc

i.e. BA+AF=BC

i.e. BC=AB+CD

Over P as PE perpendicular to AD to E, even AP

It is easy to find the congruence of EPD and CPD, and if de=dc and p is the midpoint of BC, then BP=PE can be obtained

From the edges and corners, abp and ape are congruent, and ab=ae can be deduced, so ab=ae=ad-de=ad-dc

Because c is the midpoint of AB, AC=BC, and because AD=BE, CD=CESo the triangle ACD is all equal to the triangle BCE (SSS).

So a= b

ac=cb,ad=be,cd=ce, so the triangle acd is all equal to the triangle ceb, so the angle a is equal to the angle b

Prove that two triangles, ACD and BCE congruence.

If the conditions of the question are wrong, there is no solution to the problem.

Hello, it would be much better if you gave a picture!