As shown in the figure, in the triangle ABC, point O is a moving point on the side of AC. a straight line Mn is parallel to BC through point O, and the bisector of Mn ∠ BCA is set at point E, The bisector of the outer angle of intersection ∠ BCA is at point F (1) try to explain: EO = fo; (2) when point 0 moves to where, the quadrilateral aecf is a rectangle? And explain the reason

As shown in the figure, in the triangle ABC, point O is a moving point on the side of AC. a straight line Mn is parallel to BC through point O, and the bisector of Mn ∠ BCA is set at point E, The bisector of the outer angle of intersection ∠ BCA is at point F (1) try to explain: EO = fo; (2) when point 0 moves to where, the quadrilateral aecf is a rectangle? And explain the reason

Proof: 1
∵ EC bisection ∠ BCA
∴∠ECA＝∠ECB＝∠BCA/2
∵ FC bisection ∠ ACG
∴∠FCA＝∠FCG＝∠ACG/2
∴∠ECA+∠FCA＝∠BCA/2+∠ACG/2＝（∠BCA+∠ACG）/2
∵∠BCA+∠ACG＝180
∴∠ECA+∠FCA＝180/2＝90
∴∠ECF＝∠ECA+∠FCA＝90
∵MN∥BC
∴∠OFC＝∠FCG
∴∠OFC＝∠FCA
∴OF＝OC
∵MN∥BC
∴∠OEC＝∠ECB
∴∠OEC＝∠ECA
∴OE＝OC
∴OE＝OF
2. When o is at the midpoint of AC, aecf is rectangular
∵ o is the midpoint of AC
∴AO＝CO
∵OE＝OF,∠AOE＝∠COF
The total value of Δ AOE is equal to Δ COF
∴AE＝CF
Similarly, it can be proved that AF = CE
The parallelogram aecf
∵∠ECF＝90
The rectangular aecf