Isosceles right triangle ABC, angle ABC 90 degrees, AB BC, angle BAC angle ACB 45 degrees.

solution, triangle ABC, BAC=60°

ab=6So, ac=6 cos60°=3

There is also a question that can have b ac=60° 60° 60°=120°ab =6 so, there is by the cosine theorem.

b′c|²=|ab′|²ac|²-2|ab′|×ac|×cos120°

So. b′c|=3√7

Hope it helps.

1.Equal and perpendicular, because eg and bg are the hypotenuse midlines of two right triangles with cg as the hypotenuse, and both are equal to 1 2cg, the angle egp = 2 angle acp (the outer angle is equal to the sum of the two inner angles, and because eg = gc), the angle egp = 2 angle bcp, so the angle egb = angle egp + angle bgp = 2 (angle ecp + angle bcp) = 2 angle acb = 90, so perpendicular.

2.Establish.

Summary. Both AB and AC are equal in length.

Isosceles right triangle, A angle is a right angle, BC angle is 45 degrees, BC is known to be 1170 long, find how long AB and AC are.

Both AB and AC are equal in length.

Because the angle eggplant cover A is 90 °, and the angle B is 45 °, so ab ac bc root number 2 is equipped with Nachong 1170 root number 2 ha.

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Can <> write an equation?

That's it. Can you see it.

Can be thanked. What about this root number 2.

The root number 2 is ha.

This is something to keep in mind.

Solution: Only consider as shown in the following figure, the point M is close to B, and the point N is close to C, and the order is BMNC, and the situation of BNMC is not considered

ABM rotates 90° counterclockwise around point A to ACG, and connects GN to ABM ACG

So: ab=ac

bm=cg=1

Because: abm= acg= acn=45°So: gcn= acg+ acn=90° according to the Pythagorean understanding: gn= (cn 2 + cg 2) = (3 2 + 1 2) = 10

Because: AM rotates 90° to AG

So: mag=90°= man+ ganSo: mag=45°+ gan=90°

So: mag = gan = 45°

Because: am=ag,an is common.

So: amn agn (corner edge).

So: mn=gn=10

So: mn = 10

If it's the order of the bnmc:

bn=cg=bm-mn=1-mn>0，0cm=cn-mn=3-mn>0

Because: GM2=CM2+CG2

So: mn 2=(3-mn) 2+(1-mn) 2 solution gets: mn=4- 6>1, which is not true.

To sum up, it can only be the order of BMNC.

Proof :1): DAC+D=90°, AEF+DAC=90°, so: D= AEF= BEC

acb=90°,∠bfa=90°

ac=bcso: triangle.

The ADC is all equal to the triangle BEC

So: cd=ce

2) Let BC=1=AC, and (1) gets: BE=AD, and FB is the angular bisector.

The liquid rush is the height of the bottom edge, and it must also be the middle line of the bottom annihilation on the edge of the orange.

So ad=2af, so be=2af

Hello: Is this it?

At the same time, point Q moves from point B to point C at a speed of 2 meters and seconds (when a point arrives, all stop moving.

1) Which point arrives first?

2) Let the area of the triangle ACB be Y1 and the area of the triangle QAB be Y2 after x minutes, and write the analytic expressions of Y1 and Y2 respectively about X.

3) When the moving time is in what range: The area of the triangle PCB is greater than the area of the triangle The area of the triangle PCB is less than the area of the triangle QAB?

1) The problem gives the motion time x minutes, gives the respective motion speed, can find the size of the line segment, and then use the area formula of the right triangle to write the function relationship respectively;

2) (1) know the analytic formulas of each of them, list the equations according to the equality of the area of the question, and obtain the answer by solving the equations;

3) According to the requirements of the question, list the inequalities and get the answers by solving the inequalities

Solution: (1) According to the problem: y1 = 12 pb cb = 12 (4-x) 4 = 8-2 x (0 x 2).

y2= 12bq•ab= 12×4•2x=4x（0＜x≤2）

2) When y1=y2, 8-2x=4x

x = 433) when y1 y2, 8-2x 4x

x 43 when y1 y2, 8-2x 4x

x 43Answer: (1) The functional relations are: y1=8-2x(0 x 2); y2=4x（0＜x≤2）；

2) Move at the same time for 43 minutes; These two triangles are equal in area;

3) When the moving time is 0 x 43, the area of the PCB is greater than the area of the QAB; At 43 x 2, the area of the PCB is smaller than that of the QAB

If you are satisfied, remember to adopt it!

Your praise is my motivation to move forward.

Hee-hee......

I like to do this kind of question, but you don't.

1) Proof of: abc = 90 degrees, ab = bc, a = c = 45 degrees, 1+ epf + 2 = 180 degrees, epf = 45 degrees, 1 + 2 = 135 degrees, 3 + c + 2 = 180 degrees, c = 45 degrees, 3 + 2 = 135 degrees, 1 = 3.

i.e.: ape= cfp.

2) Solution: In ABC, abc=90 degrees, ab=bc=4, from the Pythagorean theorem:

ac^2=ab^2+bc^2=32

AC=4 root number 2, p is the AC midpoint, PA=PC=2 root number 2, ape= cfp, a= c, ape cfp, CF PA=pc AE, CFXae=paxpc=8, AE=8 X, S2=s cpf

1/2)cp*cf*sinc

1 2)*(2 root number 2)x*[(root number2) 2]=x,area of s ape = (1 2)*ap*ae*sina = (1 2)*(2 root number 2)*(8 x)*[root number 2) 2]=8 x,area of s abc = (1 2)*ab*bc

8，∴ s1=s△abc-s△cpf-s△ape=8-x-8/x，y=s1/s2, y=(8-x-8/x)/x

i.e.: y=(8x-x 2-8) x 2

(1) Proof that: because the angle abc = 45 degrees, ab = bc, so angle a = angle c = 45 degrees, because angle 1 + angle epf + angle 2 = 180 degrees, angle epf = 45 degrees, so angle 1 + angle 2 = 135 degrees, because angle 3 + angle c + angle 2 = 180 degrees, angle c = 45 degrees, so angle 3 + angle 2 = 135 degrees, so angle 1 = angle 3.

i.e.: Angular ape = angular cfp.

2) Solution: 1) Because in the triangle ABC, the angle ABC=90 degrees, AB=BC=4, so from the Pythagorean theorem, it can be obtained: AC 2=AB 2+BC 2=32, AC=4 root number 2, because P is the midpoint of AC, so PA=PC=2 root number 2, because the angle APE=angle CFP, angle A=angle C, so the triangle APE is similar to the triangle CFP, so CF Pa=PC AE, CFXae=PaxPC=8, because CF=X, so AE=8 x because s2 = area of the triangle cpf = (1 2)cp*cf*sinc

1 2)*(2 root number 2)x*[(root number 2) 2]=x, area of triangle ape = (1 2)*ap*ae*sina = (1 2)*(2 root number 2)*(8 x)*[root number 2) 2] = 8 x, area of triangle abc = (1 2)*ab*bc=8, so s1 = triangle abc area - triangle cpf area - triangle ape area.

8-x-8 x, because y=s1 s2, y=(8-x-8 x) x i.e.: y=(8x-x 2-8) x 2

Picture, why didn't I see it.