Through the vertex A of the square ABCD as the line segment AE to make DC DE, cross DC to G, make DF

1. (1) ABCD is a square.

ad=dc=ab=1, adc=90°df ae dae= fdc (co-angle of the same adf) dc=de=ad

ADE is an isosceles triangle.

dae=∠dea=∠fdc

In RT DFE.

fde+∠dea=90°

i.e. fac+ cde+ dea=90°

2∠dea=90°-60°

dae=∠dea=15°

df=ad sin15°=1 ( 6- 2) 4=( 6- 2) 42).

dc=de=ad

ADE is an isosceles triangle.

df=ad×cos75°=1×（√6-√2）/4=(√6-√2)/4

2) When the CDE is close to 90°, BP > AP. The equation simply doesn't hold up.

1) Use the cosine theorem to find AE, and then use the Pythagorean theorem to find DF

Summary. Hello, I have seen your question and am sorting out the answer, please wait a while In the square ABCD, the point E is the midpoint of the CD, connecting AE, passing the point B as BF vertical AE to F, intersecting AD to H, connecting.

Hello, I have seen your question and am sorting out the answer, please wait a while In the square ABCD, the point E is the midpoint of the CD, connecting AE, passing the point B as BF vertical AE to F, intersecting AD to H, connecting.

1.The triangle bge is similar to the selling chaos and eliminating abe and bfc, so abe and bfc are similar, and because ab=bc, they are congruent 2The Pythagorean theorem yields ag=8 root in the middle of the known sign 5; gh=8/5;hd=8 root number 5, then can be found.

Proof: As eg vertical AC in G

Angle ace = 30 degrees, then eg = ec 2;

If DH is perpendicular to H, then DH=AC2

and de parallel auspicious in ac, so dh=eg, i.e., ac jin potato front 2=ec 2, ac = ec

then the angle AEC = (180 degrees - angle ace) 2 = 75 degrees;

Angle AFE = angle FAC + angle ACF = 75 degrees.

So, horn AFE = horn hand laugh AEC, AE equals af

Pass the perpendicular line of C to make ED, and cross the extension line of ED to G

Since abcd is a square, then ade=45°, cdg=45° and let the side length of the square be a

Then cg=( 2 2)a, ce=ac=( 2)a, then ecg=60°

Then we can find: ACE=90°-60°=30° AEC=1 2(180°- ACE)=75° AFE= CAD+ ACE=75°= AEC then AEF is an isosceles triangle, so AF=AE

∠a+∠ecf=180

AFC + AEC = 180

Again, let return AEC+ ced=180

afc=∠ced

and d= fbc=90, bc=dc

dec≌△fbc

Let ed be auspicious slippery god 1

ae∥bg△aef∽△bgf

bf=1,af=3,ae=1

1:3=bg:1

bg=1/3

bc=2bg:bc=1/3:2=1:6

Answer: d= fbc=90°, dc=bc, balance.

DCE ECB=90°= ECB BCF, DCE= BCF, and the dCE BCF asa ,ed=fb, let the square side length = 2, then AE=ed=FB=1, let BG=X, then get from FBG fae:

BG AE=FB FA, X 1=1 Mu Pure 1 2, X=1 3, BG BC=1 3 2=1 6

Solution: Bridge annihilation square ABCD

AB BC CB AD, agar a ABC BCD D 90 DCE + collapse mitigation ECB 90

ec⊥fc∠ecf＝90

fcb+∠ecb＝90

fcb＝∠dce

DCE is fully equal to BCF

de BFE is the midpoint of AD.

ae＝ad/2＝ab/2

df＝ab/2

af＝ab+df＝3ab/2

bc∥adbf/af＝bg/ae

ab/2）/（3ab/2）＝bg/（ab/2）bg/ab＝1/6

bg/bc＝1/6

bg：bc=1：6

Let bc=1, and solve bf=, bg=1 3 from similar triangles, so bg:bc=1:6