Mathematics winter vacation homework questions in the second year of junior high school, math applic

1.(1) BCE congruent AFE

Proof: AD BC

f=∠bce，∠b=eaf

and e is the midpoint of the imitation wide ab.

AE=BE, BCE, Quanqinliang, et al. AFE

Solution: (2) ab=6

E is the midpoint of AB.

ae=be=3

AB BC again

In the RT Bridge BCE, it is obtained according to the Pythagorean theorem.

ec = 16 + 9 = 5 under the root number

and BCE congruent AFE

ec=ef=5ef=5

Let the l1 function be y=kx and replace a(4,3) with people.

3=4kk=3/4

y=3/4x

The x-axis of point A is point C

AC X-axis.

oc=4，ac=3

In RT AOC, it is obtained according to the Pythagorean theorem.

oc=5 and oa=ob

Point b is (0,-5).

Let the l2 function be y=kx+b, and substitute a(4,3) and b(0,-5 into it. 3=4k+b①

5=b Substituting .

3=4k-5

k=2y=2x-5

Hopefully oh, I'm typing slowly, haha.

I'll help you! Because the triangle ADE can coincide with ABF, the triangle AFB is all equal to the triangle ADE, that is, the angle EAD = angle BAF (the congruent triangle corresponds to the same angle), that is, AF=AE (the congruent triangle corresponds to the sides equal) Because the angle BAD is 90° (the inner angle of the square is 90°), it should be the angle BAD = angle BAE + angle EAD (known), so the angle BAD = angle FAE = angle BAE + angle EAD = angle BAF + angle BAE = 90° (equal substitution) so the triangle AEF is an isosceles right triangle.

Be patient, I hope it helps.

First, you first have to prove that the two congruent triangles are equal to each other

AD+DE=AE, AB+BF =AF, so AE>DE, AF>BF, and because AD=AB, AE=AF, DE=BF

Second, prove that they are right triangles.

The triangle ecf is a right triangle, so ef = ec +bc +bf, and because af = ab +bf , ae = ad +de , so ef -af -ae = ec -ad -de <0,, the question is wrong.