Jtdylktuy

In parallelogram ABCD, points E and F are chosen on sides AB and CD, respectively, so that AE = DE
and CF/DF=2/3. Find the ratio of the area of triangle BFC to the area of quadrilateral BEDF.

I'd appreciate it if someone could show a step-by-step solution to the problem. :)

Choosing these points is in general not possible:
If $AE=DE$, then $E$ has to be on the perpendicular bisector of $A$ and $D$. But $E$ is also on $AB$. Hence $E$ is on the intersection of this line and this segment. But if the segment $AD$ is too short, then there is no intersection.

In fact, if $alpha$ is the angle at $A$, then we need $ABge frac{AD} {2 cos(alpha)}$.

In parallelogram $ABCD$, given $AE=DE$ and $frac{CF}{DF}=frac{2}{3}$, then joining $DB$, and taking $AD$ and $angle DAB$ as fixed, and sliding $AD$ to the right until $E$ coincides with $B$, then$$frac{triangle BFC}{BEDF}=frac{2}{3}$$since triangles under the same height have areas proportional to their bases.

On the other hand, if we slide $AD$ increasingly to the left, quadrilateral $BEDF$ becomes an ever greater fraction of the lengthening trapezoid $ABFD$. Disregarding $triangle ADE$ as negligible, then, and since a parallelogram is double a triangle of equal base and height, the ratio of $triangle BFC$ to quadrilateral $BEDF$, as $AD$ moves to the left, approaches the ratio $frac{triangle BFC}{ABCD}$, i.e. diminishes indefinitely toward$$frac{2}{3}cdotfrac{1}{2}=frac{1}{3}$$

I am not confident this answer meets the real intent of OP's problem, but based on the information given, the most I can conclude is that$$frac{1}{3}<frac{triangle BFC}{BEDF}<frac{2}{3}$$