As shown in the figure, in △ ABC & nbsp;, point O is a moving point on the edge of AC, and a straight line Mn ‖ BC is made through point O. suppose that the angular bisector of Mn intersecting ∠ BCA is at point E, and the outer angular bisector of intersecting ∠ BCA is at point F. (1) try to explain EO = fo; (2) when point O moves to where, is the quadrilateral aecf rectangular? (3) when the point O moves to where, and what condition does △ ABC satisfy, the quadrilateral aecf is a square? And explain the reason

As shown in the figure, in △ ABC & nbsp;, point O is a moving point on the edge of AC, and a straight line Mn ‖ BC is made through point O. suppose that the angular bisector of Mn intersecting ∠ BCA is at point E, and the outer angular bisector of intersecting ∠ BCA is at point F. (1) try to explain EO = fo; (2) when point O moves to where, is the quadrilateral aecf rectangular? (3) when the point O moves to where, and what condition does △ ABC satisfy, the quadrilateral aecf is a square? And explain the reason

(1) ∵ Mn ‖ BC, ∵ OEC = ∵ BCE, ∵ OFC = ∵ GCF, ∵ CE bisecting ∵ BCO, CF bisecting ∵ GCO, ∵ OCE = ∵ BCE, ∵ OCF = ∵ GCF, ∵ OCE = ∵ OEC, ∵ OCF = ∵ OFC, ∵ EO = Co, fo = Co, ∵ EO = fo. (2) when point O moves to the midpoint of AC, quadrilateral aecf is rectangular (3) when the point O moves to the midpoint of AC, and △ ABC satisfies the condition that ∠ ACB is a right triangle, the quadrilateral aecf is a square. It is known from (2) that when the point O moves to the midpoint of AC, the quadrilateral aecf is a rectangle, and Mn ‖ BC is known. When ∠ ACB = 90 °, then ∠ AOF = ∠ COE = ∠ COF = ∠ AOE = 90 °, AC ⊥ EF, aecf is square