OA=OB=4, the slope of straight line AB is tg( 180-45)=- 1, and the equation y=-x+4.

(2) The two triangles are similar. If X=EM, the weights of the two triangles are equal, and X=EM=OM/2=√2.

(3) I left before dinner.

Keep coming.

Area of OBDC:

Lower base OB=4, upper base ED=CD-CE=y-CE=-x+4-x, height x.

S 1=(4-x+4-x)x/2=x(4-x)

S2= triangle OAM area-triangle OCE area

=OA*2/2-x^2/2=4-x^2/2

s 1+s2=x(4-x)+4-x^2/2=(-3/2)x^2+4x+4

This is a parabola.

Because the quadratic coefficient is negative, the parabola opens downward, from the maximum value to the maximum value.

The maximum value is

(4ac-b^2)/4a=20/3