In the quadrilateral ABCD, ab = CD, e and F are the midpoint of BC and ad. the extension lines of Ba and EF intersect at m, the extension lines of CD and EF intersect at n, proving that angle ame is equal to angle dne

In the quadrilateral ABCD, ab = CD, e and F are the midpoint of BC and ad. the extension lines of Ba and EF intersect at m, the extension lines of CD and EF intersect at n, proving that angle ame is equal to angle dne

Hint, the detailed process of their own supplement
Connect BD, take the midpoint g, connect FG, eg
FG is parallel to AB, FG = AB / 2, angle GFE = angle ame
Eg parallel CD, eg = CD / 2, angle FEG = angle dne
FG=EG
Angle GFE = angle FEG
Angle ame = angle dne

In the quadrilateral ABCD, ab = CD, e and F are the extension Ba of the midpoint of AD and BC, respectively, and the intersection n of CD and the extension line of EF is at MF, so ∠ ame = ∠ DNC is proved

As shown in the figure, in the quadrilateral ABCD, e is the midpoint of AD, CE intersects Ba extension line at point F, and CE = EF. (1) try to explain: CD ‖ AB; (2) if be ⊥ CF, try to explain: CF bisects ∠ BCD

It is proved that: (1) e is the midpoint of AD, ∵ de = AE, in △ Dec and △ AEF, de = AE, ∵ Dec ≌ AEF = Fe, ≌ Dec ≌ AEF (SAS), ∵ d = EDF, ∥ CD ∥ AB; (2) ce = EF, be ⊥ CF, ∥ BC = BF, ∥ FCB = f, ≌ Dec ≌ AEF, ≌ DCE = f, ∥ DCE = FCB, ∥ CF bisection ∥ BCD

As shown in the figure, in the parallelogram ABCD, e is the midpoint of AB, f is the point on ad, EF intersects AC with G, AF = 2cm, DF = 4cm, Ag = 3cm, then the length of AC is ()
A. 9cmB. 14cmC. 15cmD. 18cm

∵ quadrilateral ABCD is a parallelogram, ∵ BC = ad = 6cm, BC ∥ ad. ∵ EAF = ∠ EBH, ∵ AFE = ∠ bhe, AE = be, ≌ AFE ≌ bhe, ≌ BH = AF = 2cm. ∵ BC ∥ ad, ∥ agcg = afch, that is 3cg = 28, then CG = 12, then AC = Ag + CG = 15 (CM)