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