(1) extends BC, with point F as its vertical line and point H as its crossing point.

Because BAE+BEA = 90 degrees, FEC+EFH = 90 degrees, BEA+FEC = 90 degrees.

So ∠BAE=∠FEC(2) Because ∠BAE=∠HEF, ∠AEB=∠BE=FH, AE=EF, so △ ABE △ EHF, so BE=HF, AB=EH, AB=BC. So ∠FCH=45 degrees because ∠FCH+∠FCD=90 degrees, so ∠FCD=45 degrees (3) because the quadrilateral ABCD is a square, ∠ABC=90 degrees, ∠BAC=∠BCA=45. So ∠BAE+∠CEG=90 degrees ∠CEG+∠EGC=90 degrees ∠ BAE = ∠ CEG ∠ AEB = ∠ EGC so △ AB = EC△ ECG (.