In the parallelogram ABCD, is the midpoint of e-type CD, a point on the edge of F-type BC, and the angle FAE = angle ead, EF and AE perpendicular? Give reasons

In the parallelogram ABCD, is the midpoint of e-type CD, a point on the edge of F-type BC, and the angle FAE = angle ead, EF and AE perpendicular? Give reasons

Conclusion: EF ⊥ AE
Prolonged AE, BC to h, easy to syndrome △ ade ≌ △ HCE
∴AE=EH,∠H=∠DAE
∵ angle FAE = angle EAD
∴∠H=∠FAE
∴FA=FH
∵AE=EH
∴EF⊥AE

In ▱ ABCD, the bisector BC of ∠ bad intersects at point E, and the bisector DC intersects at point F. if ∠ ABC = 120 ° FG ‖ CE, FG = CE, connect dB, DG, BG respectively, the size of ∠ BDG is ()
A. 30°B. 45°C. 60°D. 75°

Extend the intersection of AB and FG at h, connect HD. ∵ ad ∥ GF, ab ∥ DF, ∵ quadrilateral ahfd is parallelogram, ∵ - ABC = 120 °, AF bisects ∵ bad, ∵ DAF = 30 °, ∵ ADC = 120 °, ∵ DFA = 30 °, ∵ DAF is isosceles triangle, ∵ ad = DF, ∵ parallelogram ahfd is diamond, ∵ ADH