(20 15 & amp； #8226; As shown in the figure, in △ABC, ∠ ACB = 90, D and E are the midpoint of BC and BA respectively, and d E and F are connected on the extension line of DE, and AF = AE.

(1) verification: quadrilateral ACEF is a parallelogram;

(2) If the quadrilateral ACEF is a diamond, find the degree of ∠ b. 。

analyse

(1) CE=AE=BE You can get AF=CE according to the fact that the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse, and then you can get ∠ 1=∠2 according to the properties of the three lines of an isosceles triangle on a line, and then you can get ∠F=∠3 according to the equilateral angles, and then you can get.

(2) According to the fact that all four sides of the rhombus are equal, we can get AC=CE = AE, so that △AEC is an equilateral triangle, and then we can get ∠ CAE = 60 according to the fact that every angle of the equilateral triangle is 60, and then the two acute angles of the right triangle are complementary angles.

The solution (1) proves that ∠ ACB = 90, e is the midpoint of BA,

∴CE=AE=BE，

AF = AE，

∴AF=CE，

In △BEC, BE = CE, d is the midpoint of BC,

∴ED is the midline of the bottom of isosceles △BEC,

∴ED is also the bisector of isosceles vertex angle △BEC,

∴∠ 1=∠2,

AF = AE，

∴∠F=∠3，

∵∠ 1=∠3,

∴∠2=∠F，

∴CE∥AF，

CE = AF，

∴ Quadrilateral ACEF is a parallelogram;

(2) solution: ∵ quadrilateral ACEF is a diamond,

∴AC=CE，

From (1), AE=CE,

∴AC=CE=AE，

∴△AEC is an equilateral triangle,

∴∠CAE=60，

In Rt△ABC, ∠ b = 90-∠ CAE = 90-60 = 30.

This topic examines the nature of rhombus, the judgment of parallelogram, and the judgment and nature of equilateral triangle. The midline on the hypotenuse of right triangle is equal to half of the hypotenuse, and the two acute angles of right triangle complement each other. Remembering all the properties and judging methods is the key to solving the problem.

C. How can I see the original picture after submission? This is a waste of time. I thought I was looking for a topic, hehe.