A proof question

Intercept af=ad on ab and connect fe because ae bisects dab so fae is equal to angular fad.

aef adf triangle congruence.

Therefore, EF=DF, be, bisecting ABC, can prove that FBE congruence, cbe, fe=ce

Therefore de=ce i.e. e is the midpoint of DC.

2 is also intercepted af=ad to connect fe because ae bisects dab so fae is equal to angular fad.

aef adf triangle congruence hence ef=df and e is dc midpoint hence de=ce

fb=ab-af=ab-ad=bc be=be so fbe congruent cbe fbe = angular cbe thus be bisected abc

Pass E as EF BC and AB to F

AE bisector angle DAB, BE bisector angle ABC

So ead= eab= aef ebc= eba= bef, so af=fe=bf i.e. f is ab midpoint.

So EF is the median of trapezoidal ABCD, so E is the DC midpoint.

If E is the midpoint of DC, it is easy to know that EF is the median line of trapezoidal ABCD and F is the midpoint of AB.

Because ae bisects the angle dab, ead= eaf= aef, af=ef=bf

So feb= fbe= ebc

So be bisected by ABC

The condition of AD+BC=AB does not seem to be used.

Good guys, good guys.

I didn't dare to test this thing for ten years.

I counted it and failed.

But the equation is certainly true.

Initial attempts.

<> then it won't be spicy.

The speed of light fails, because simplification can't go on.

Looked at the image.

Periodic function. The conclusion does hold.

If you want to go on with it.

Either the trick is (the strength).

Or simply engage in cyclicality (dog head).

Good guys, what ignorance.

The proof is as follows, 9=20 degrees 18=10 degrees.

Converting to an angle value means that it is easier to type than an radian representation.

So -2tan( 9)sin( 18)+tan( 9)=tan( 18)+2tan( 18)sin( 18).

Suppose the length of rectangular ABCD is ab=cd=x and the width of AD=CB=EF=y, then because rectangular ABCD is similar to rectangular BCEF

So ef dc=ce cb

i.e. y x = ce y

And because the quadrilateral AFED is a square.

So ef=de=x-ce=y

So ce=x-y

i.e. y x=(x-y) y

Sorted out x 2-xy-y 2=0

The root finding formula yields x1=(y+ 5y) 2 or x2=(y- 5y) 2 (rounded).

So x y=((y+ 5y) 2) y=( 5-1) 2 so the rectangular abcd is ** rectangle.

Let DF and AC intersect at point H.

Then FHA= CHD

Because FHA+ F= CHD+ C=90

So f= c

According to ASA, the congruence of two triangles can be obtained, ac=ef

Rotate around the line cd, bc is a circle with a radius of 10, area = 10 * 10 * = 100

ab is a cylinder with a height of 15 and a radius of 10, area = 10 * 2 * * 15 = 300 ad = root number (10 2 + 5 2) = root number 125, radius 10 cone surface, area = root number 125 * 10* = 50 root number 5

Surface area = sum of the above items = (400 + 50 root number 5).