The quadrilateral ABCD is made up of two congruent isosceles right triangles

<> connect AC to extend PH to AD and connect FG

GPCD is rectangular and has GD=PC

In the isosceles right-angled triangle dgh, f is the midpoint on the hypotenuse.

There is gd df = root number 2, on which it has been proven that gd = pc

There is pc df = root number 2

ac is diagonal, ac ad = root number 2

pc df = root number 2, ac ad = root number 2, acp = adf = 45° so acp adf

There is pac= fad

PAF = PAC + CAF = FAD + CAF = 45°PH AB with be EH = AB PH

pb ad, there is be ed=pb ad, i.e., ed be=ad pb and pb=ph, ab=ad

So be eh=ed be

And df = fh, eh = ef-fh, ed = ef + df so be 2 = (ef-df) * (ef + df) i.e. be 2 + df 2 = ef 2

Connect the AC to prove that the triangle PAC and FAD are similar.

So pac= fad

This gives paf=45

The triangle ABE and PHE are similar.

ph=bp=1-x bh=√2 (1-x)be/bh=ab/(ab+ph)

be=√2 (1-x)/(1+1-x)=√2 (1-x)/(2-x)df=√2 x/2

ef=√2-be-fd

Simplification and squaring can be demonstrated to be 2 + df 2 = ef 2

According to the formula of the sum of the inner angles of the polygon = (n-2) 180 degrees n=4, that is, the sum of the inner angles of the quadrilateral is = 360 The three angles of the equilateral briefs are 60 degrees, that is, one of the inner angles of the quadrilateral is 60 degrees, and because of an isosceles triangle with a top angle of 120 degrees, the other two angles are 30 degrees. If a silver source is 120 degrees top.

(1) Proof that in the parallelogram ABCD, ab cd, ad bc, ead= f, baf= e

and ead= baf, e= f

ce=cf．

i.e. CEF is an isosceles triangle

2) Solution: In CEF, the sum of CE and CF is exactly equal to the circumference of the parallelogram Prove as follows: from (1) we get ead= f= baf= e, de=ad, ab=bf

ce cf=cd ad cb ab．

That is, the sum of the perimeters of the parallelogram is equal to the sum of CE and CF

It can be seen from Qi that the four Zen points of ABCD are round, and AB is the diameter of cracked dust.

And the quadrilateral ABCD is an isosceles trapezoid.

Point E at **?

1) Quadrilateral.

ABCD is a liquid shape for the following reasons: ABC and ACD are two congruent equilateral triangles.

ab=bc=ac, ad=cd=ac, ab=bc=cd=ad, quadrilateral abcd is a diamond; (2) Be=CF The reasons are as follows: abc and acd are two congruent isopremature triangles, ab=ac, b= acd=60°, the property of the rotation of the finger to the object, bae= caf=40°, in abe and acf, b= acd=60° ab=ac bae= caf , abe acf(asa), be=cf; (3) As shown in the figure, be=cf is justified as follows: abc and acd are two congruent equilateral triangles, ab=ac, b= acd=60°, by the nature of rotation, bae= caf, in abe and acf, b= acd=60° ab=ac bae= caf , abe acf(asa), be=cf