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A Mathematical Geometry Problem of Grade Eight in Junior Middle School
1, △ADC and △AED are congruent (angular bisector, right angle, male side), so CD=ED, ∠CDA=∠EDA.

2. see parallel DE. So △CFD and △EFD are congruent (public side DF, ∠CDA=∠EDA, CD=ED).

3. Therefore, CD=DE=EF=FC, that is, the quadrilateral CDEF is a diamond.

4. Proof of proposition

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