Urgent!! The second geometry will do to solve

Question 1: Because AB and CD intersect at the point E, the angle AED = angle CEB, and because EA = EC, ed = EB, the triangle AED is all equal to the triangle CEB (corner edge).

Question 3: Because C is the midpoint of AB, AC=BC, and because CD is parallel to BE, the angle ACD=angle CBE, and because CD=BE, the triangle ACD is all equal to the triangle CBE (corner edge).

A: ae=ce, ed=eb, aed= cebδaed δceb(sas).

Two: Because c is the midpoint of AB, AC=BC

And because cd is parallel to be, the angle acd = angle cbe, and because cd = be, the triangle acd is all equal to the triangle cbe (corner edge).

1.Proof: ae=ce, ed=eb, aed= ceb (equal to apex angles).

According to the corner theorem, δaed δceb

3.Proof: Point c is the midpoint of AB, ac=cb

cd∥be,cd=be.

acd=∟cbe

According to the corner theorem, δacd δcbe

Subtraction is all this: add the opposite of this number!

That's the concept. Examples:

Got it?

1、bf=ce

Proof: Connecting AF, investigating ABF and AEF, AB=AD=AE, the common edge AF determines the congruence of two triangles from the hypotenuse right-angled edge.

So bf = ef, and because it is easy to know that cef is an isosceles right triangle, fe = ce

So bf=ce

2. Equal. The parallel line of the point E is parallel to the AD intersection AB at the point F, because AD is parallel to BC, so ADEFBC is parallel to each other, and because E is the midpoint of CD and divides CD, F is also the midpoint of AB and divides AB; And because ab bc, which is parallel to fe, ab ef, so ef is the perpendicular bisector of ab.

So ae=be

Connecting AF is due to being a square.

ad=ab,ae=ad

yes, ae=ab

Angle AEF = Angle ABF = 90°

af=af gives aef the full equals the triangle abf

bf=ef takes the midpoint f of ab and connects EF

AD parallel BC, AB vertical BC

e, f are the midpoints of each waist of the trapezoid.

The EF is parallel to the AD

EF is perpendicular to AB

, EF is the mid-perpendicular line of AB.

Abe is an isosceles triangle, ae=be

1.Connecting AF, it can be proved that the triangle AEB and ABF are congruent, and BF EF is known

The triangle cef is an isosceles right-angled triangle, so ce ef, so bf ce

2。equal, passing the point e as a parallel line to AD, can be proved according to the median line theorem.

1.Let the side length of the square be x, ae=ab=bc=cd=da=x, ac= 2x, ef=ce=( 2-1)x, cf= 2ef=(2- 2)x bf=x-(2- 2)x=( 2-1)x bf=ef (this is equivalent to a calculation problem).

2.Take the middle point F of AB and connect EF

E is the CD midpoint.

ef∥ad∥bc

ab⊥bcab⊥ef

ae=af (the high and midlines coincide in an isosceles triangle).

de is the perpendicular bisector of bc, be=cethen ae + be = ae + ce = ac = 8

So: ab = perimeter of abe - (ae + be) = 14 - 8 = 6

When the point E is on the outside of a square ABCD, it is obtained from ABCD being a square and ADE being an equilateral triangle.

cde＝90°＋60°＝150°，de＝ad＝dc，∠dec＝∠ecd＝（180°－150°）÷2＝15°

The same can be done for AEB 15°

then bec aed aeb dec 60° 15° 15° 30°

ad=dc=db

ADC and DCB are isosceles triangles.

again b=30

dcb=30,∠cdb=120

adc=60

acb=90，∠dcb=30

acd=60

The sum of the inner angles of the triangle is equal to 180

A=60 ACD is an equilateral triangle.