ABC and CED are both isosceles right triangles, A CDE 90, point D on the side BA, DE intersects BC a

1)、be⊥bc。

2) ABC and CED are isosceles right triangles, A= CDE=90°, CD=2, DCA=30°

ad=1,ac=√3

ab=ac=√3

In BCE, CE=2 2

be=√2，bc=√6

s△bce=1/2×√6×√2=√3

s△adc=1/2×√3×1=√3/2，s△abc=1/2×√3×√3=3/2，s△cde=1/2×2×2=2

s△dbe=s△cde+s△bce-s△cdes△abc-s△acd+s△bce-s△cde

Proof that: (1) ABC is an isosceles right triangle, b=45°, 1+ 2=135°

and def is an isosceles right triangle, 3=45°

2=∠4，∠b=∠c=45°，△bem∽△cne；

2) Same as (1) BEM CNE, BECN

emne again be=ec,eccn

emne∴ec

em cnne, again ecn= men=45°, ecn men

The original question should be that the ECD is an isosceles right triangle.

Correct solution: CB 2 in an isosceles right triangle ACB

ac 2=ab 2 cb=ac and ab=ad+db=5+12=17 so 2bc 2=17 2 solution : bc=17 2 2 in a triangle cdb according to the cosine theorem cd 2=db 2+bc 2-2*db*bccos cad=12 2+(17 2 2) 2-2*12*17 2 2*cos45° solution cd=13 2 2 2 in an isosceles right triangle ecd cd 2+ec 2=de 2 so de 2=2cd 2=2*( 13 2 2) 2=13 2 so de= 13 2=13

13 ADCs can be rotated 90° clockwise.

Solution: Isosceles right-angle ABC, ECD is known

So AC=BC,EC=DC, ECD= ACB=90°, i.e.: ACE= BCD (the co-angles of the same angle are equal) (just look at the figure) i.e.: ACEs congruent BCD(SAS).

You're in good hands:

I didn't think of a simple way to solve this problem for the time being, and I thought about it a lot, so I could only solve it, and the process is as follows:

If a is the high ah on the side of bc, it is easy to know ah=bh=ch, and let bae= , then it is easy to get the eh= , eah= 4- , then.

bebh-eh

ah-ah*tan(π/4-α)

ah-ah[(1-tanα)/1+tanα)]ah*[2tanα/(1+tanα)]cfch-ahtanα

ah(1-tanα),ef

ah*[tanα+tan(π/4-α)

ah*(1+tan )1+tan ),be +cf ah ah [(1+tan )1+tan ) ah*(1+tan )1+tan )]ef , i.e. be +cf =ef, prove!

I'm on my phone.,The square symbol can be displayed normally.,It's proof that it's a little bland.。。

If you find a better way, welcome to talk!

If you still have questions, feel free to ask them again! Thank you!

Let's just use the Pythagorean theorem.

This may be extreme, but it can be done.

Analysis: This question can be answered as a moving point question.

Proof: In the triangle ABC, let the point E coincide with the point B'The angle BAC = 90 degrees, and the triangular liquid deficit ABC is an isosceles triangle. , angle b = angle c = 45 degrees.

When E coincides with point B, since the angle AEF = Angle Eaf, Sobi Tsai uses the angle AFE = 90 degrees that is, AF is perpendicular to the point F

Since the triangle ABC is an isosceles triangle.

So if AF bisects BC perpendicularly, then EF=CF,'Point e coincides with point b. ,be=be²=0

Again','ef=cf,cf²＝ef²,be²＋cf²＝ef²

I have seen this kind of solution in the Olympiad problem, but I don't often regret burying it, and I can't remember the general method.

de=4 shows that DF=2 root number 2 is greater than AC AD distance and is 2 root number 2

Suppose the coincident part dd'The distance is a a square + a square = x square y = 1 2 times a square.

Substitution yields y=1 4 times x square (x is greater than 0 and less than 2 root number 2.)

This is too troublesome, you upload the function diagram above the question, and I'll solve it.

From the meaning of the title, it can be seen:

The function image is a parabola with the opening facing down with respect to x=2 symmetry.

Because abc is isosceles, it can be deduced that ac=bc=1

Let y=(general).

When y=1 (maximum), x=2

So the negative b is divided by 2a=2, so b=-4a, c=0 substituting y=1 when x=2 gives a= , b=1 so the function relation is: y=squared + x