AB / / CD, ∠ DAB = ∠ BCD, try to explain AD / / CD

AB / / CD, ∠ DAB = ∠ BCD, try to explain AD / / CD

Because ab ∥ CD, ∠ B + ∠ C = 180 ° and ∠ a = ∠ C
So, ∠ a + ∠ C = 180 ° so ad ‖ CB

As shown in the figure, the known angle AB is parallel to AC, AB is equal to AC, ad is parallel to AE, ad is equal to Ae
As shown in the figure, AB is perpendicular to AC, AB equals AC, ad is perpendicular to AE, ad equals AE. Try to explain that be is equal to CD and be is perpendicular to CD.

Zinckd "EF parallel AB, DF cross AC at point F" is "DF parallel AB, DF cross AC at point F"?
AE and CF are not necessarily equal
reason:
Because ad bisects ∠ BAC
Therefore, bad = CAD
Because de ∥ AC, DF ∥ ab
So the quadrilateral AEDF is a parallelogram,
And ∠ bad = ∠ ADF
Therefore, ADF = CAD
So AF = DF
So Quad AEDF is diamond
So AE = AF
Are AE and AF equal
If AE = CF
Then AF = CF, that is, D is the midpoint of AC
So D should also be the midpoint of BC
The triangle ABC should be isosceles triangle
There is no such condition in the original title
So AE and CF are not necessarily equal
For reference! Jswycfgs
two thousand one hundred and twenty-one

In trapezoidal ABCD, ab ‖ CD, ∠ a = 90 °, ab = 2, BC = 3, CD = 1, e is the midpoint of AD

It is proved that the extension line of extending CE intersection Ba is at point G, that is, the intersection point is g, ∵ e is the midpoint of AD, ∵ AE = ed, ∵ ab ∥ CD, ∵ CDE = ∵ gae, ≌ DCE = ≌ age, ≌ CED ≌ GEA, ≁ CE = Ge, Ag = DC, ≁ GB = BC = 3, ≁ EB ⊥ EC