Oh gosh... In which grade are you ? Because I've found a solution, but it's a bit tricky ~ !
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 ><
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...
(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°.
By Pythagorean theorem,
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 !
I made a little mistake : the area of BEC is
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
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 ?