1 is correct: take f as the vertex to make a vertical line, let the height of triangle CEF be FH, then the area of triangle CE×FH=Xc×Yc as the vertex to make a vertical line, and let the height of triangle EFD be EM, then the area of triangle is EM×DF=Xd×Yd.

Because Xc×Yc=Xd×Yd=k, the areas of the two triangles are equal.

2 is right: from 1, we can see that the areas of triangle CEF and triangle EFD are equal.

Let the vertical line of c image EF be p, and the vertical line of d image EF be q.

CEF area of triangle =CP×EF EFD area of triangle = =DQ×EF

So CP = dq

So CD is parallel to EF, so the two triangles are similar.

4 is right: according to the solution of 2, CD is parallel to EF.

Because AF is parallel to the CE quadrilateral and AC=EF is a parallelogram, AC = EF.

Similarly, the quadrilateral BEFD is a parallelogram, so BD = ef.

So AC = BD