As shown in the figure, in ▱ ABCD, e and F are two points on AC, be ∥ DF

As shown in the figure, in ▱ ABCD, e and F are two points on AC, be ∥ DF

It is proved that: as shown in the figure, ∵ quadrilateral ABCD is a parallelogram, ∵ ad ∥ BC, ad = BC, ∵ DAF = ∠ BCE. Also ∵ be ≌ DF, ∵ be = DF. In ∵ be ∥ DF, ∵ quadrilateral BEDF is a parallelogram

In the parallelogram ABCD, e and F are two points on AC respectively, and be is perpendicular to AC at e, DF is perpendicular to AC at F

Because e and F are two points on AC respectively, and be is perpendicular to AC and DF is perpendicular to AC and F
So be is parallel to DF (two lines perpendicular to the same line are parallel)
Because the quadrilateral ABCD is a parallelogram
So AB is parallel and equal to CD
So BAE = DCF
And because angle AEB = angle DFC = 90 degrees
So the triangle AEB is equal to the triangle DFC
So be = DF
So the quadrilateral BEDF is a parallelogram

As shown in the figure, in the parallelogram ABCD, e is the midpoint of Ba, DF is perpendicular to BC, and the perpendicular foot is F. then is the angle ade = the angle BFE?
 

Make a straight line eg ∥ BC through e, intersect DF with G in parallelogram ABCD 1, ∵ ad ∥ BC (the opposite side of parallelogram is parallel), eg ∥ BC (done) ∥ ad ∥ BC ∥ eg (a straight line is parallel to another straight line, also parallel to its parallel line) ∥ ade = ∥ DEG, ∥ BFE = ∥ FEG (the parallel line intersects the third straight line