The point on the AD side that is away from point A X is connected with the point on the BC side that is away from point B Y. ..

Then the original trapezoid becomes two quadrangles, with the graphic area S 1 on the left and the graphic area S2 on the right.

The whole trapezoidal area S=2*(2+4)/2=6.

S 1=(X+Y)*2/2=X+Y

S2=[(4-X)+(2-Y)]*2/2=6-(X+Y)

S 1=S2

X+Y=6-(X+Y)

X+Y =3

It can be seen that there is no need to pass through the fixed point. As long as the sum of upper x and lower y =3, the requirements can be met.