0
$\begingroup$

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. :)

$\endgroup$
  • $\begingroup$ Did you make a typo somewhere, please check! $\endgroup$ – Oldboy Nov 16 '18 at 15:02
  • 1
    $\begingroup$ What have you tried? Where are you stuck? $\endgroup$ – Jens Nov 16 '18 at 15:26
0
$\begingroup$

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.

enter image description here

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

$\endgroup$
0
$\begingroup$

ratio of triangle to quadrilateral in a parallelogram 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$$\frac{1}{3}<\frac{\triangle BFC}{BEDF}<\frac{2}{3}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.