24. (The full score of this small question is 14) Solution: (1) It is easy to prove Δ ABF ≌ΔADH, so AF=AH (2) As shown in the figure, rotate Δ ADH 90 degrees clockwise around point A, as shown in the figure, it is easy to prove Δ AFH ≌ΔAFM, and get FH. It is easy to get BG = 1-X, BF = 1-Y, FG = X+Y- 1. (1-x) 2+(1-y) 2 = (x+y) from Pythagorean theorem. Therefore, the area of the rectangular EPHD is 0.5.25. (The full score of this small question is 14 points) Solution: (1)OC= 1, so q=- 1, from which we can know that the area is 0.5OC×AB=, and get AB=, let a (a). 0, so p=. Therefore, the analytical formula is: (2) Let y=0 and solve the equation, so a (0) and b (2 2,0) can be obtained in the right triangle AOC, and AC= can also be obtained as BC=. Obviously, the triangle ABC is a right triangle. AB is the hypotenuse, so the diameter of the circumscribed circle is AB=, so. (3) Existence, AC⊥BC,① If AC is the base number, BD//AC, the easy-to-find analytical formula of AC is y=-2x- 1, and the analytical formula of BD can be set to y=-2x+b, b (2,0 0) D (0 0,9) ② If BC is the base number,