# urgent simple geometry proof (equilateral triangle)

• May 22nd 2008, 11:38 PM
cbbplanet
urgent simple geometry proof (equilateral triangle)
ABC is an equilateral triangle with sides of 2 cm. BC is extended its own length to point D and point E is the midpoint of AB. ED meets AC at point F. Find the area of quadrilateral BEFC in square centimeters in simplest radical form.

I've drawn a figure and everything and tired to proof with a 30-60-90 triangle theorem but nothing is working!! please show me how to solve this using a 2 column proof.

THANK YOU SO MUCH!! really need it before tomorrow morning ><
• May 23rd 2008, 01:13 AM
Moo
Hello,

Oh gosh... In which grade are you ? Because I've found a solution, but it's a bit tricky ~ !
• May 23rd 2008, 01:15 AM
cbbplanet
^ really? my teacher said it was supposedly to be easy by drawing altitudes and finding the centroid and using the ratio of 1:2 on the lengths...

ps. im in 10th (is that good or bad?)
• May 23rd 2008, 01:18 AM
Moo
Quote:

Originally Posted by cbbplanet
^ really? my teacher said it was supposedly to be easy by drawing altitudes and finding the centroid and using the ratio of 1:2 on the lengths...

ps. im in 10th (is that good or bad?)

I don't know... I just hope you know much about the ratio theorem :)
I'll post it rapidly, I have lessons in 20 minutes :(

This what it looks like for me
• May 23rd 2008, 01:23 AM
cbbplanet
hm, thanks! i think i get it a little better now. let me try to work it out or at least BS it ><

thanks though! picture helped
• May 23rd 2008, 01:32 AM
Moo
Ok, first of all, let's set the plot.

C is the midpoint of BD (you said it was extended its own length).
E is the midpoint of AB.

Hence in triangle ABD, we can conclude that EC and AD are parallel because EC is the midpoints line.

Plus, CE is a median of the triangle ABC. But because it's equilateral, it's also an altitude. Therefore, there is a right angle in the red part on E. Furthermore, there is also one in A (because they are parallel).

H and I are such that EH and FI are perpendicular to BC. This will help for like the latter step of the demonstration... but the most difficult in my opinion...

----------------------------

You want the area of BEFC.
This is equal to the area of BEC and EFC.

I think it looks evident to you how much is the area of BEC, because it's a right angle triangle...

$A_{BEC}=\frac{\sqrt{5}}{2}$ (this has been mixed with an application of the Pythagorean theorem, I'll let you check it if you're not sure).

-------------------
Now, let's calculate EH.

We know that angle ABC is 60°.
Therefore, $\sin 60=\frac{EH}{BE}$

$\sin 60=\frac{\sqrt{3}}{2}$
And BE=1.

--> $\boxed{EH=\frac{\sqrt{3}}{2}}$

By Pythagorean theorem, $BH=\frac 12$

Try to find FI, then you will have the area of CFD (FIxCD/2). After that, it's all about substracting areas to others...

I'm REALLY and sincerely sorry !!!! But I'm in a big hurry :'(
I just hope it will help you to go through the problem !
• May 23rd 2008, 04:42 AM
Moo
Back ~

I made a little mistake : the area of BEC is $\frac{\sqrt{3}}2$

Now, look at triangle ABD. E is the midpoint of AB and C is the midpoint of BD. Therefore, the interesection of DE and AC, F, is the barycenter of the triangle. So we have $\frac{DF}{DE}=\frac 23$

Now, consider the triangle BED, with parallel lines EI and FH. You can apply the Thales theorem, with proportion 2/3. This will give you FI...

Do you understand what it yields ?
• May 23rd 2008, 07:25 AM
cbbplanet
yay!! i think so. thank you so much!!