1. Connect AN, DN, ∵ ∠ BA, cf =∠BDC = 90°, ∴AN=DN (the midline on the hypotenuse of a right triangle is equal to half of the hypotenuse), and the point where the distance from ∴ Mn ⊥ ad to both sides of the angle is equal is at the middle vertical line of the angle. ∵AD//BC,∴△AFQ∽△BCQ。 ∵F is the midpoint of AD, ∴ Similarity ratio k= 1:2, AB=AQ means that A is the midpoint of BQ. It is easy to prove that △BCE≌CDF(SAS) ∠ DCF = ∠ CBE ∴∠ BPQ = 90. ∴ AP = AB = AQ3。 If the crossing points E are EP//AB, EQ//BC, AE=BP, DE=CQ, ∠B=∠EPQ, ∠C=∠EQP, ∴∠peq = 90°, F is the midpoint of PQ. ∴ef= 1/2pq= 1/2(bc-ad)4。 If the crossing point D is DE //AC, the quadrilateral is a parallelogram, and AD=CE=3. ∴bd⊥de,be= 10. ∵ac⊥bd and ∵AB=CD, AD//BC, ∴△ Abd. AC=BD=DE。 ∴△BDE is an isosceles Rt triangle S= 1/ 2× 10×5=25 5. ∫BD is the diagonal of square ABCD, ∴AF=CF,△AFD≌△CFD(SAS)∠DAF=∠DCF. Yizheng △ Abe △ DCE (SAS) +∠ AEB = ∠ DCF.