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Mathematics in the second day of junior high school ~ ~ Find the answer
(1) As shown in figure 1, in the square ABCD, points E and F are points on the sides of DC and BC respectively, and satisfy ∠ EAF = 45, connect EF, and verify that DE+BF = ef.

Feel how to solve the problem and complete the following blanks:

Rotate △ADE clockwise by 90 degrees around point A to get △ABG. At this time, AB and AD overlap, which can be obtained by rotation:

AB=AD,BG=DE,∠ 1=∠2,∠ABG=∠D=90,

∴∠ABG+∠ABF=90 +90 = 180,

Therefore, point G, point B and point F are on the same straight line.

≈EAF = 45 ∴∠2+∠3=∠bad-∠eaf=90-45 = 45。

∵∠ 1=∠2,∴∠ 1+∠3=45 .

That is ∠GAF =∞.

FAE

.

And AG=AE, AF=AF.

∴△GAF≌

△ EAF

.

girlfriend

=EF, so de+BF = ef.

(2) Use the accumulated experience and knowledge in (1) solution to solve the following problems:

As shown in Figure 2, in the right-angled trapezoidal ABCD, AD ∥ BC (AD > BC), ∠D = 90°, AD=CD= 10, E is a point on the CD, and ∠BAE = 45°, DE=4. Find the length of BE.

(3) The proof idea of analogy (1) completes the following problems: put two isosceles right triangles ABC and AFG together in the same plane, where A is the common vertex and ∠ BAC = ∠ AGF = 90. If △ABC is fixed, △AFG rotates around point A, AF and AFG.

Decomposition solution: (1) According to equivalent substitution, ∠GAF=∠FAE.

The △ GAF △ electric arc furnace is obtained by SAS.

∴GF=EF,

So the answer is: FAE;; ; △ electric arc furnace; GF;

(2) A is AG⊥BC, and the extension line of CB is at G 。

In the right-angled trapezoidal ABCD,

∵AD∥BC,∴∠C=∠D=90,

∠ CGA = 90,AD=CD,

∴ Quadrilateral AGCD is a square.

∴CG=AD= 10.

Given BAE = 45,

According to (1), be = GB+de.

Let BE=x, then BG=x-4,

∴BC= 14-x.

In Rt△BCE, ∫BE2 = BC2+CE2, that is, x2 = (14-x) 2+62.

To understand this equation, we get: x=

58

seven

.

∴BE=

58

seven

(3) Proof: As shown in the figure, rotate △ACE 90 degrees clockwise around point A to the position of △ABH.

Then CE=HB, AE=AH, ∠ abh = ∠ c = 45, and the rotation angle ∠ eah = 90.

Connect HD, in △EAD and △HAD,

AE = AH,∠HAD=∠EAH-∠FAG=45 =∠EAD,AD=AD。

∴△EAD≌△HAD,

∴DH=DE,

And < hbd = < abh+< Abd = 90,

∴BD2+HB2=DH2,

That is BD2+Ce2 = DE2 ... (11)

(1) Get ∠GAF=∠FAE by equivalent substitution between angles, then get △ GAF △ EAF by SAS, and get the answer;

(2) Let A BE AG⊥BC, pass BC and continue to G, then CG=AD= 10 can be obtained from the property of square, and then the length of BE can be obtained by Pythagorean theorem and equation;

(3) The equation is proved by using the rotation property and Pythagorean theorem.