In triangle ABC, because the angle ABC=90 degrees. So triangle ABC is a right triangle. EF vertical BC; So the angle BDE=90 degrees. So AC is parallel to DF. Point d is the midpoint of BC, which is derived from the meaning of the question. So BD=DC, then DE is the bit line in the right triangle ABC. So point e is the midpoint of BA. Then CE is the center line of the hypotenuse of the right triangle, so BE=AE=CE. And because CF=AE and CF=CE=BE=AE. In triangle FDC and triangle FDCFD=FD, angle FDC= angle FDB, DC=DB. So two triangles are congruent. So FC=FB. In quadrilateral BECF, CF=CE=BE=BF. So the quadrangle BECF is a diamond.

2。 . . . . When the angle α is equal to 45 degrees, the quadrilateral is a square. In the triangle CEA, CE=AE is obtained from 1. So the triangle CEA is an isosceles triangle. Because angle A is equal to 45 degrees, angle ECA is equal to 45 degrees. Equal sides and angles. The angle BEC equals the angle A plus the angle ECA, which equals 90 degrees. The outer angle of a triangle is equal to the sum of two non-adjacent inner angles. The quadrilateral BECF obtained from 1 is a diamond, because the BEC angle is equal to 90 degrees. So the quadrilateral is a square at this time.

Hehehe. . . . . . . . . . . . .