In the quadrilateral ABCD, point E is the midpoint of BC, point F is the midpoint of CD, and AE is perpendicular to AB, AF ⊥ CD, connecting EF to prove AB = ad When AB, CD and BC are related, the △ AEF is an equilateral triangle

In the quadrilateral ABCD, point E is the midpoint of BC, point F is the midpoint of CD, and AE is perpendicular to AB, AF ⊥ CD, connecting EF to prove AB = ad When AB, CD and BC are related, the △ AEF is an equilateral triangle

[correction: the title should be AE ⊥ BC]
prove:
Connect AC
∵ AE ⊥ BC, e is the midpoint of BC
The AE is the vertical bisector of BC
Ψ AB = AC [the distance from the point on the vertical bisector to both ends of the line segment is equal]
⊙ AF ⊥ CD, f is the midpoint of CD
The AF is the vertical bisector of CD
∴AC=AD
∴AB=AD.①
If ⊿ AEF is an equilateral triangle
Then ∠ AEF = ∠ AFE = 60 & # 186;
∴∠CEF=∠CFE=30º
∴CE=CF
∴BC=CD
∵CE=CF,AE=AF,AC=AC
∴⊿AEC≌⊿AFC（SSS）
∴∠EAC=∠FAC=30º
∴∠BAE=∠CAE=30º
∴∠B=60º
Then ⊿ ABC is an isosceles triangle
∴AB=BC=CD.②