In the isosceles triangle ABC, D is the midpoint of BC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively （1）△BDE≌△CDF (2) When ∠ A is equal to what degree, Quad AEDF is square? Prove your conclusion Point D is on BC, and a is the vertex angle

In the isosceles triangle ABC, D is the midpoint of BC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively （1）△BDE≌△CDF (2) When ∠ A is equal to what degree, Quad AEDF is square? Prove your conclusion Point D is on BC, and a is the vertex angle

It is proved that: (1) in the isosceles △ ABC, if ∠ A is the vertex angle, then AB = AC, D is the midpoint of BC, de ⊥ AB, DF ⊥ AC, the perpendicular feet are e, F, ≁ de = DF respectively, (the distance from the midpoint on the bottom edge of isosceles triangle to the two waists is equal), and DB = DC (known) ≌ RT △ BDE ≌ RT △ CDF, (the oblique and straight angle sides correspond to two equal right triangle congruence)
(2) When ∠ A is equal to 90 degrees, the quad AEDF is square
∵ Ca ⊥ AB, de ⊥ AB, ∥ de ∥ Ca, similarly, DF ∥ AB, ∥ quadrilateral AEDF is a parallelogram (two groups of parallelograms with opposite sides parallel to each other are parallelograms). When ∠ A is equal to 90 degrees, quadrilateral AEDF is a rectangle (parallelograms with right angles at one corner are rectangles), and then de = DF, (proved) ∥ rectangular AEDF is a square (a group of rectangles with equal adjacent sides are squares)