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Math Help - urgent simple geometry proof (equilateral triangle)

  1. #1
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    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 ><
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  2. #2
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    Hello,

    Oh gosh... In which grade are you ? Because I've found a solution, but it's a bit tricky ~ !
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    ^ 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?)
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    Quote Originally Posted by cbbplanet View Post
    ^ 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
    Attached Thumbnails Attached Thumbnails urgent simple geometry proof (equilateral triangle)-equilat.jpg  
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  5. #5
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    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
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  6. #6
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    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 !
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  7. #7
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    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 ?
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  8. #8
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    yay!! i think so. thank you so much!!
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