A geometric problem with affection, a plane geometry problem

Proof: ca=co

am=mocm⊥ab

In a right-angled triangle BMC.

pb=pcpm=1 2bc (the right triangle hypotenuse is the middle line equal to half the length of the hypotenuse) The same is true. In a right-angled triangle BNC.

pn=1/2bc

pm=pn

You can get 2 points, and you will get bounty points and 20 points

Geometry problems with affection? What do you mean?

Connect to the IO, IP, and ID

Easy Evidence POI DOI

The radius of the circumscribed circle of the doi is the same as the radius of the circumscribed circle of the poi, and op 6, pio 90° 1 2 pqo 135° is the diameter of the circumscribed circle of the pio, so the poc is an isosceles right triangle pc 2op 6 2

Thus the radius of the doi is 3 2 choose d

You do it right, turn the original diagram into a square, you can find ae as two times the root number six, the diagonal is twice the root number of the side length, ac = four times the root number two.

（1）∵bc‖do

Angular CBA = Angular DOA

and d= bac

Angular CBA+ D = Angular DOA+ BAC

That is, angular bac = angular dao = 90°

Ad is the tangent of a semicircle o.

2) 3 roots of 10

(1) d= bac, doa= abc, acb=90°, dao=90°, proven.

2) From (1), we know that the two large triangles are similar, and both are similar to AEO, BC:OE = AE:AC, AE:AD=OE:OA, and it is easy to know AE=3, OE=1, OA= 10 from the known conditions

ae:ad=oe:oa,，3：ad=1：√10，ad=3√10

1、∵od‖bc，∴∠doa=∠cba

d= bac, bca= dao, and abc doa and bca=90°, dao=90°, so ad is the tangent of the semicircle o.

2. According to the similarity theorem, know abc aeo

oa ab = ae ac=1 2, get ac = 6 ab = ac2 + bc2 = 2 10, oa = 10, bc oa = ac ad, ad=3 10

∵bc//od∴∠cba=∠doa

and d= bac dao= bca=90° ad is the tangent of the semicircle o.

bc ac=ao ad=ab 2ad ae oe=ac bc oe ao=ae ad solution equation.

Solution: (1) Proof of: d= bac

Whereas, BC OD then CBA= AOD

abc∽△ doa

And acb is the angle of the diameter of the semicircle o, then acb=90° then acb= doa=90°, i.e., ad ab

Ad is the tangent of a semicircle o.

2) BC OD is easy to know AOE ABC

Therefore, ae ac=ao ab=1 2 gives ac=2ce=6ao=1 2 (|bc|²+ac|) = 10 from (1) abc doa

ad/ac=ao/bc

Get ad=3 10

Obviously, the area of the triangle dam is 1 4

Obviously, the triangular EBM is similar to the triangular EDC, and the chain pin MB=CD2, so the area of the triangular EDC is 4 times that of the triangular EMB.

Let the area of the triangle EMB be X, then the area of the triangle EDC is 4X, and the area of the triangle DMB is 1 4, and the area of the triangle MCD is 1 2, and let the surface of the triangle MDE be Y, then:

x+y=1/4, 4x+ y = 1/2

The solution is x=1 12 y= 1 6

The triangle MBC area is 1 4, so the triangle EBC area = 1 4 - 1 12 = 1 6 and the last black part area is 1 6 + 1 6 = 1 3

Before I took it, there were 8 that took A, B, and C, and after that, there were still only 8.

8 There are 3 faces on each corner of each corner for a total of 8 corners.

Because :,ad bc, the isotope angles are equal, b= ead=30° bad=180- ead=150°

AD is the bisector of EAC, EAC=2* EAD=60°

Angular BAC = 180- EAC = 120°

Angle c = 180-30-120 = 30°

∠ead=∠b=30°

eac=2∠ead=60°

Angle BAC = 180 degrees - Angle EAC = 120 degrees.

Angle c = 180 degrees, minus angle B, subtracted angle BAC is equal to 30 degrees.

The angle bad is equal to the angle BAC + the angle DAC is equal to 150 degrees. or angle BAD = 180 degrees minus angle EAD = 150 degrees.

I hope you can understand the Chinese characters in the back.

∵∠b+∠bac+∠c=180°

again a=90°

c=180°-90°-30°=60°

eac=90°

eac=90°

ead=90°÷2=45°

dac=45° ,bad=90°+45°=135° and then modify it.

Because AD is parallel to BC, the angle EAD is 30, BAD is 150, EAC is 60, and C is 30