Analysis: (1) It is known that the coordinates of point A are (4,0) and the coordinates of point C are (1, 2). According to the "two-point method", the analytical formula of linear AC can be obtained. (2) Let B be BH⊥OA in H, and the coordinates of point B can be obtained according to the properties of isosceles trapezoid. The analytical formula of straight line AC can represent the line segment PQ, given AM, and then find the area of △AMQ, and find the maximum value according to the properties of quadratic function. (3) When △AMQ is an isosceles triangle with MQ as the waist, there are two cases: ①QM=QA, ②QM=MA, which can be solved according to graphic characteristics and Pythagorean theorem.

Solution: Solution (1) Let the analytical formula of straight line AC be y=kx+b, and substitute point A (4,0) and point C (1 2) into 14k+b = 02k+b = 2. The solution is k =-2/3b = 8. According to the property of isosceles trapezoid ∴AH= 1, then op = OA-ah-HP = 4-1-bn = 3-t: point q is point PQ =-2/3 (3-t)+8/3 on AC ∴. PQ = 1/2(4-2t)(2/3t+2/3)=-2/3t？ +2/3t+4/3； When t= 1 /2, s max =3/ 2.

(3) There are two situations: ①QM=QA, and MP=AP, that is, 3-3t=t+ 1, t=0.5(2 points) ②QM=MA, that is, QM2=MA2, MP2+PQ2=MA2 is obtained from Pythagorean theorem, that is, (3-3t). +(2 /3 t+2 /3)？ =(4-2t)？ , t 1=59/ 49, t2=- 1 (truncated) ∴ When t=0.5 or t 1=59 49, △AMQ is an isosceles triangle with MQ as its waist.

This topic examines the solution of analytical formula of straight line, the representation of triangle area in coordinate system, the maximum value of quadratic function, and the conditions for finding isosceles triangle.