The first question, (1) perimeter is unchanged. The circumference of the quadrilateral OCMD is 2 (MD+MC), and MD and MC are the vertical and horizontal values of point M, that is, the circumference is 2 (X+Y), that is, it is known that Y =-X+4, x+y=4, and the circumference is always 8. (2) area s = xy = x * (4-x) =-(x-2)? +4, because 0 < x < 4, so when x=2, point M is the midpoint of AB, and the maximum value is 4. (3) When OCMD is square, X = Y = 2, the area is 4, and its diagonal CD equation is Y =-x=y=2. When the square moves to the right, y=-x+a+2 and ABy=-x+4 are always parallel. When the translation moves to the right, it is obviously a piecewise function. The intersection of diagonal y=-x+a+2 and AB is its node, that is, when A = 20 < A ≤ 2, the overlapping part is the subtraction of two isosceles RT△ areas, and S =1/2 * 2-1/2 * A = 2-/kloc-0. When 2 < a < 4, the overlapping part is isosceles RT△, and S =1/2 (4-a) (4-a) =1/2a? -4a+8 Question 2 1)OE=OF

EO=CO because EF∨BC, CE and CF are angular bisectors, and ∠CEF=∠BCE=∠ACE.

Similarly, CFE = FCD = ACF, so OF=OC, so OE=OF. 2) ∠ EF > CF =1/2 ∠ ACB+1/2 ∠ ACD = rt, so ef > cf and BCFE can't be rhombic. 3) When AECF is a square, obviously O is the midpoint of EF, so O is the center of the square and O is the midpoint of AC.

At this time, it is enough that AECF is a parallelogram. Because ∠ECF=RT, it is rectangular.

It is required that EC=FC meets the square condition, that is, ∠ CEF = 45, ∠ACB=RT O is moved to the midpoint of communication, △ABC is RT△ABC, ∠ACB=RT is enough.