# In parallelogram ABCD, points E and F are chosen on sides AB and CD, respectively.

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. :)

• Did you make a typo somewhere, please check! – Oldboy Nov 16 '18 at 15:02
• What have you tried? Where are you stuck? – Jens Nov 16 '18 at 15:26

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 $$AB\ge \frac{AD} {2 \cos(\alpha)}$$.
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$$\frac{2}{3}\cdot\frac{1}{2}=\frac{1}{3}$$
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$$\frac{1}{3}<\frac{\triangle BFC}{BEDF}<\frac{2}{3}$$