Come and solve this geometry problem!! Thank you for helping me solve a geometry problem!!

Note: The square of b below is expressed as b*b

In the same way, we get be*be=(a*a+c*c) 2-b*b 4 2 formula cf*cf=(a*a+b*b) 2-c*c 4 3 formula.

1 + 2 + 3 to get ad*ad+be*be+cf*cf=3*(a*a+b*b+c*c) 4

That is, you write "ad2+be2+cf2" equal to the sum of the squares of the three sides of the triangle and multiply it by 3/4.

Well, that's it.

Make AG parallel BC, so that AG=BC, connect GC, make up the graph into a parallelogram, because E is the midpoint of AC, AC is the diagonal of the parallelogram ABCG, so BG is 2 times of BE, according to the sum of squares of the diagonal of the parallelogram is equal to the sum of squares of each side, BG2+AC2=AB2+BC2+CG2=GA2, we can calculate BG2=1 4(2C2+2A2-B2), and the same can be obtained AD2=1 4(2C2+2B2-A2), cf2=1 4(2b2+2a2-c2), so ad2+be2+cf2=3 4(a2+b2+c2).

It is a triangle that waits for it.

In a right-angled triangle CFA, the angle CFE = angle BCA minus the angle CAF = 90 degrees - angle CAF

In a right triangle AED, the angle AF=Angle EDA-Angle DAE=90 degrees-Angle DAE, because the angle CAF=Angle DAE, and the angle AF=Angle FEC, so the angle CFE=Angle FEC

So the triangle cef is an isosceles triangle.

From the definition of a circle, the center of the circle is on the bisector of the angle cab, which is set to am:x Shixun 3+y 8=:x 6+y 4=1.

Coupling, unraveling: x=2, y=4 3, the circle is (2,4 3), the distance to ac:x=0 is 2, so the radius of the circle is 2.

Solution: Let o and ab and ac be tangent to d and e respectively, then there is do=oe, because c=90°, so the quadrilateral ceod is a square, (draw your own picture).

Let cd=x, then ce=do=x

When the point p is moved for 2 seconds, cp = 2 * 2 = 4 cm

Because in the three-hu cluster angular PCB, the DO is parallel to the CB, so

do:cb=pd:pc

i.e.: pants closed cherry blossom x:6=(4-x):4

x=A:, The radius of the state O is.

The first floor is right. The 2nd floor is not right, the CEOD is not graphic, E is on the AC, not on the BC.

I can't do it with a bury to defend the front liquid.

ab bisecting virtual kai dc] is incorrect, and it is changed to [ab ping pure reputation scum line dc] to prove: de bisecting adc

cde=1 2 do quiet adc

BF bisected ABC

2=1/2∠abc

by abc=adc

2=∠cde

AB DC again

1=∠cde

Do CE vertical AB in E, CF vertical AD in F

AC squared bad

ce=cfbc=cd

Right triangle, hl

cbe==cdf

Angular FCD = BFE

a+∠ecf=180

Angular FCD = BFE, BCD = FCE

bcd+∠bad=180°

∠1+∠2=2∠a

Connecting aa, it can be proved that the triangle is equal to the sum of two non-adjacent inner angles.