∴CE=AE=BE

AF = AE

∴AF=CE

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

∴ED is the center line of the bottom edge of isosceles △BEC.

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

∴∠ 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.

Extended data:

Analysis:

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

(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.

This topic examines the nature of rhombus, the judgment of parallelogram and 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 are complementary angles. Memorizing all the properties and judgment methods is the key to solving problems.