In △ ABC, point P is a moving point on edge AC, and a straight line Mn ∥ BC is made through point P. let Mn intersect ∥ BCA and its outer angle bisector be at point E, and intersect ∥ BCA and its outer angle be at point F (1) Verification: PE = PF (2) Can the quadrilateral BCFE be a diamond when point P moves on edge AC? Explain why (3) If there is a point P on the side of AC, let the quadrilateral aecf be a square, and AP / BC = (√ 3) / 2

In △ ABC, point P is a moving point on edge AC, and a straight line Mn ∥ BC is made through point P. let Mn intersect ∥ BCA and its outer angle bisector be at point E, and intersect ∥ BCA and its outer angle be at point F (1) Verification: PE = PF (2) Can the quadrilateral BCFE be a diamond when point P moves on edge AC? Explain why (3) If there is a point P on the side of AC, let the quadrilateral aecf be a square, and AP / BC = (√ 3) / 2

(1) ∵ CE bisection ∠ BCA,
∴∠BCE=∠ECP,
And ∵ Mn ∥ BC,
∴∠BCE=∠CEP,
∴∠ECP=∠CEP,
∴PE=PC；
Similarly, PF = PC,
∴PE=PF；
(2) When point P moves to the midpoint of AC side, the quadrilateral aecf is a rectangle
According to (1), PE = PF,
∵ P is the midpoint of AC,
∴AP=PC,
A quadrilateral aecf is a parallelogram
∵ CE and CF are equally divided into ∠ BCA and ∠ ACD,
And ∠ BCA + ∠ ACD = 180 °,
∴∠ECF=∠ECP+∠PCF= 1/2（∠BCA+∠ACD）= 1/2×180°=90°,
The parallelogram aecf is a rectangle;
(3) It is proved that if the quadrilateral aecf is a square, then AC ⊥ EF, AC = 2AP
∵EF‖BC,
∴AC⊥BC,
The ∧ ABC is a right triangle, and ∠ ACB = 90 °,
∴cos∠A= AC:BC=2AP:BC= √3,
∴∠A=30°．