# Parallelograms in triangles

if posssible, could you only give me a few theorems in order to assist me in this question. Thankyou in advance! Links to simple websites would also be appreciated.

In triangle $ABC$ $F$ is midpoint of $AC$ $E$ is midpoint of $AB$

$X$ and $Y$ are points along $CB$

If $EX$ and $FY$ are parallel and $EFYX$ is a parallelogram, what is the ratio of the area of $EFYX$ to $ABC$?

All I understand is that similarity or congruency should somehow be used.

Show/prove/use the following:

$\;FE\;$ is a midsegment of $\;\Delta ABC\;$ and thus $\;\begin{cases}(1)&FE//BC\\{}\\(2)&|FE|=\frac12|BC|\end{cases}\;$

Not only that: the midsegment $\;FE\;$ bisects any segment from vertex $\;A\;$ to side $\;BC\;$ , so:

$$Area_{EFYX}=\frac{|EF|\cdot h}2\;\;,\;\;h=\text{height of paralellogram}\;EFYX$$

$$Area_{\Delta ABC}=\frac{BC\cdot H}2\;\;,\;\;H=\text{height to \;BC\; in triangle}\;\;\Delta ABC$$

But, as mentioned above, $\;h=\frac H2\;$ , so...

Stewart's Theorem and the Law of Cosines are useful for this type of problem. Here is a helpful link:

http://www.artofproblemsolving.com/Wiki/index.php/Stewart's_Theorem

Also Menelaus' Theorem will be helpful:

http://en.wikipedia.org/wiki/Menelaus'_theorem

Good luck!

• Your link to Menelaus...is a beautiful and nice link to the historical/mythological Menelaus: no mathematics there as far as I could see... – DonAntonio Sep 21 '13 at 15:42
• The link's now fine...but I can't see how this helps: in Menelaus Theorem the segment is required to intersect the three sides of the triangle and here $\;FE\;$ is parallel to $\,BC\;$ , so... – DonAntonio Sep 21 '13 at 15:45