A math problem about triangles, a math problem about triangles

Let the waist length be x, from the inscription: one part is 2cm longer than the other, 8+x 2=x+x 2+2 or 8+x 2=x+x 2-2

x=6 or x=10

Verify that you can form a triangle.

The waist length of the triangle is 6cm or 10cm.

One part is 2 cm longer than the other, indicating that the bottom is 2

The waist length of this triangle is 3cm

That's what I thought.

I don't know if it's right or not.

This is a type of problem about right triangles.

If the angle ACB = 90 degrees. Also on the hypotenuse. Then these two points can also be on it.

And the angle pcq = 45 degrees. Then there must be a little bit of the two points on the two points. Then the point p or q is on the middle line of the right triangle.

Then p or q must coincide. then the square of AP is 0So the square of PQ = the square of AP + the square of BQ.

Proof : P is on the angular bisector of DBC.

The distance from P to DB is equal to the distance from P to BC.

That is, the distance from P to AB is equal to the distance from P to BC.

In the same way, the distance from P to AB is equal to the distance from P to BC, and the distance from P to AB is equal to the distance from P to AC.

p is on the bisector of the bac.

Proof: AE is the bisector of BAC.

bae = ∠cae

AD is perpendicular to BC

DAC + C = 90 degrees, DAB + B = 90 degrees. 1）

dac = ∠cae + dae

dab = ∠bae - dae

Substituting (1) CAE + DAE + C = 90 degrees .2) bae - dae + b = 90 degrees .3) (2) -3) De.

cae + dae+ ∠c） -bae - dae+ ∠b）=0

And bae = cae, simplified.

dae=1/2（∠b-∠c)

The figure is too lazy to draw, B is an obtuse angle, D is on the CB extension, and E is on the side of BC.

Angle EAD=1 2*Angle BAC+Angle Bad=1 2*(180-Angle B-Angle C)+(Angle B-90)=1 2*(Angle B-Angle C).

Even CF. Let the area of the triangle CDF be X, and the area of CEF be Y (followed by the area) In ACD: X+3Y=1 (i.e., 1 3 of ABC) In BCE: 3X+Y=1 (i.e., 1 3 of ABC) 4X+4Y=2, and S quadrilateral DCEF=X+Y=1 2

Let s efd=s, obviously s afb=9s;

Because s ced=3*(1 3) 2=1 3, the trapezoidal area aedb=3-1 3=8 3=9s+s edb+s eda-s=;

And because s edb=s eda=2 3*1 3*s abc=2 3;

So the trapezoidal area aedb=8s+2*2 3=8 3;So 8s = 4 3; s=1/6;

So the area of the quadrilateral DCEF is s+s ced=1 3+1 6=1 2

The edge AC folds along the AD and coincides with the AE.

The corners of the two triangles are equal, and the triangles are congruent.

acd≌△aed

ae=ac=6

be=ab-ae=10-6=4

The bed is a right-angled triangle.

bed∽△bca

be/bc=de/ac

de=4×6/8=3

acd≌△aed

cd=ed=3

Fold past, triangle ACD and triangle AC'd are two identical triangles, hence ac'=ac=6, so bc'=10-6=4, set cd=c'd=x, then bc'=8-x, at this point you can build the equation, I won't say much about it, just look at the sketch I made......

Connecting de, then de is the median line, so de ac can be obtained from the two internal wrong angles corresponding to the equal.

The triangle deg is similar to the triangle ACG

So ge gc=gd ga=de ac=1 2, so ge ce=gd ad=1 3

Rotate the triangle AEC 180 degrees around the point E, and then translate it so that E and D coincide, C and B coincide, and A to A'location. Link AA'.Extend AD Sex BA'F, ab ac ab a'b=ab+bf+a'f>af+a'f=ad+df+a'f>ad+a'd=ad+ae