As shown in the figure, in the right triangle ABC, e is the intersection of the bisectors of two acute angles, ed ⊥ BC is D, EF ⊥ AC is F. can you explain that the quadrilateral cdef is a square?

As shown in the figure, in the right triangle ABC, e is the intersection of the bisectors of two acute angles, ed ⊥ BC is D, EF ⊥ AC is F. can you explain that the quadrilateral cdef is a square?

First, it is possible only when C is the vertex of a right angle
Let Ge ⊥ BC be g, then by the angular bisector theorem, EF = eg = ed
Because AC ⊥ BC, know DC ⊥ FC, and by ED ⊥ BC, EF ⊥ AC, know quadrangle cdef is rectangle
From the previous EF = ed, we know that this is a square
PS: in junior high school, the teacher said that the following points should be paid attention to: 1. If the three inner angles are right angles, they are rectangles; 2. There are two ways to prove a square: one is to prove a rectangle, and then a square; the other is to prove a diamond, and then a square