It is proved that the intersection point E makes EF perpendicular to BC, and BC intersects with F, because E is the midpoint of AD, EF is perpendicular to BC, and AD = 2ab, so AE=EF, and because the quadrilateral ABCD is rectangular, the angle A is 90 degrees, because the angle AEM plus the angle MEF equals 90 degrees, and the angle MEF plus the angle NEF equals 90 degrees, so the angle AEM equals the angle FEN, so the triangle AEM equals the triangle FEN.