First of all, 1 is right (AD = AB, AP=AE), and the included angles are equal. Let DAP be the angle 1. If BAE is angle 2, then the angle 1 = angle 2, so AEDB 4 is round, so 3 is right. Then, if AF=PQ is perpendicular to FB, we can easily know the rectangle APQF, then AF = PQ and FB = X, then we can easily know that Fe = X, so (x+1) square+(x-1) square = 5, and X=2 is the square root of 6. So 2 is wrong. So EP = root number 2, EB = root number 3, and angle peb = 90 degrees, so S △ APD+S △ APB = the square root of SAE BP = 0.5+26, so 4 is wrong. S squared ABCD = AB squared = AF squared +FB squared = the square root of 4+6, so 5 pairs.