The square problem in the second year of junior high school, and the square problem in mathematics i

<> do PK vertical BC and cross BC to K point.

It can be proved that PK is the median line of trapezoidal ACFE.

So PC=PF triple solution PCF is an isosceles triangle.

pk=1/2(bf+bc)

ck=1/2(bc-bf)

pk = root number 3ck

bf = (root number 3-1)ck

bc = (root number 3+1)ck

bf bc = 2 - root number 3

1) Isosceles. If PK AC crosses BC to K, then PK BC, and P is the midpoint of AE, then K is the midpoint of CF.

PK is both the height on the bottom edge of CF in PCF and the midline.

then PCF is isosceles.

2) Let bf=x, bc=a, then pk=(a+x) 2 Since pcf=60°, then ck=pk 3= 3(a+x) 6, i.e. cf= 3(a+x) 3

3(a+x) 3+x=a is obtained by cf+bf=bc, and (3+ 3)x=(3- 3)a

x/a=2-√3

Isosceles triangle.

pg||AC crosses CB at point G

ac||pg||EF and point P is the AE midpoint.

Point G is the midpoint of CF.

and PG CF (Easy Proof).

PG bisects CF perpendicularly

cp=pf

(1) If you calculate the angle, you can get that ade is an isosceles triangle with d as the vertex, so de=ad=16cm

2) Connect the PP'，∠pbp'= abc=90°, so pbp'is an isosceles right triangle, so pp'=3√2

3) Let the side length of the square be x, then ce=x 2, and the Pythagorean theorem gives x 2+(x 2) 2=5 2

So the area is x 2 = 125

4) There should be a picture for this question.

Typing math symbols on the computer,It's particularly annoying.,You also have to process.。。 Forget it.. I'm lazy.

Because it is a square, ABO is an isosceles right triangle, so the ratio of the three sides is 1:1: the root number is 2, so the circumference is 18x (2+1 under the root number).

Dust extension friend cd to m, so that dm=be, connected am

Prove that the Abe of the Shen Brother Jane Triangle is fully equal to ADM

AE=AM,FM=BE+DF,Angle BAE=Angle DAFAngle AFD=Angle EAF+Angle BAE= Angle FAD+ Angle BAE= Angle FAD+ Angle DamAngle FAM=Angle FAD+Angle DAM

So angular AFD = angular wide pants FAM

So fm=am

So ae=be+df

Auxiliary line: Extend CB to G so that BG=DF

The square abcd,ab=ad,ab bc,ad dc type abg is all equal to adf

gab=∠fad, ∠agb=∠afd

AF bisector EAD

gab=∠fae

ab cd kay slip afd= fab

gab =∠fab=∠gae

AEG is isosceles, AE=eg=BE+BG=BE+DF

In BEF and AHE.

bc=abcf=ae

bf = ae and bef and ahe are both right triangles.

bef≌△ahe

The same is true for ef=eh, which can be proved: hg=gf=ef=eh

bef≌△ahe

efb=∠aeh

efb+bef=90°

aeh+bef=90°

i.e.: feh=90°

Again: efg = fgh = ghe = 90°

EFGH is square.

Not equal.

Let the side length be 2a

Similar triangles are used to correspond to the proportionality of the sides and the Pythagorean theorem.

Calculate DM = 5a and mn = 13a

∵ab=ac

AH bisects the BAC

bah=∠cah

Bah and gam are on top of cah and mae are on top bah= gam cah = mae bah= cah

gam=∠mae

ag=ae ∠gam=∠mae am=am∴△agm=△ame

em=mg

Rotate the triangle APB 90° clockwise around point B to the BOC to connect the OQ ABCD to a square.

ab, bc coincide.

Easy to get OBP is 90°

bpq=90°

The PQ OB reproves that the PQOB is a square.

pq=pb