Very difficult, someone may give better solution, here is mine (

1) from C draw a line CX1 to DA making DX1=BX, from D draw a line DY1 to AB making AY1=CY.
2) Asumme four intersets of AX and BY, CX1 and BY, DY1 and CX1, AX and DY1 are B1, C1, D1, A1.
3) you can prove AX=CX1 ( using congruent triangles, ABX and CDx1), XC=AX1, so AX || CX1 (parallel rectangle, AXCX1). Same to prove BY||DY1. That makes right rectangle A1B1C1D1.
4) AB1^2 + BB1^2 = BC1^2 + CC1^2 (using right triangles with same  hypotenuse).
5) BB1/BC1 = (BX/BC = AY1/AB) = AA1/AB1 => AB1*BB1 = BC1*AA1  =>
   AB1*BB1 = BC1*CC1    (AA1=CC1 from another pair of congruent triangles)

6) Combine AB1^2 + BB1^2 = BC1^2 + CC1^2 and AB1*BB1 = BC1*CC1 =>
   AB1=BC1, and BB1=CC1  ( using sqrt plus, sqrt minutes equations)
7) That make triangle ABB1 and BCC1 congruent, so <BAX=<CBY  which make tiangle BAX right triangle or ABCD is a square.
8) One more step to prove AX=BY (congruent triangles ABX and BCY)

所有跟帖： 

• 与PhD无关,不过不是小孩子的题目,需要耐心 -flagsix- ♂ (0 bytes) () 12/06/2012 postreply 12:54:30

• 真正高手，逻辑完全正确 -桃花流水杳然去- ♀ (177 bytes) () 12/06/2012 postreply 13:07:11

• 多谢, step 4), 5) and 6) 不容易想得到, -flagsix- ♂ (49 bytes) () 12/06/2012 postreply 13:14:25

• 回复：多谢, step 4), 5) and 6) 不容易想得到, -桃花流水杳然去- ♀ (75 bytes) () 12/06/2012 postreply 13:20:03

• 我也是把图画来画去,用了好长时间. -flagsix- ♂ (0 bytes) () 12/06/2012 postreply 13:28:03

• 这个证明有漏洞 -DeanNg- ♂ (592 bytes) () 12/06/2012 postreply 18:18:45

• 每一步都没有非常仔细写，这个题整个证明比较长， -flagsix- ♂ (482 bytes) () 12/06/2012 postreply 18:51:24