1. ## Equilateral last

ABC is an equilateral triangle.
|AE|=|EC|=3 units
|DC|=2|BD|
CF is perpendicular to DE

Find |AF|=X=?

2. ## Re: Equilateral last

Show some work?

3. ## Re: Equilateral last

My work is a mess.
AE=3
BD=2
DC=4
I drew a parallel to AB from D. I tried congruent triangles but couldn't find anything. I need a clue.

4. ## Re: Equilateral last

Use Law of Cosines:
$|ED|^2 = 3^2+4^2-2\cdot 3\cdot 4 \cos(60^\circ) = 13$
So, $|ED| = \sqrt{13}$

Next, use Law of Sines:
$\dfrac{\sin 60^\circ}{\sqrt{13}} = \dfrac{\sin \left(\angle DEC\right)}{4}$

$\angle DEC = \arcsin\left(\dfrac{2 \sqrt{39}}{13}\right)$

That gives
$\angle ECF = 90^\circ-\arcsin\left(\dfrac{2 \sqrt{39}}{13}\right)$
$\angle AFC = 120^\circ - \angle CEF = 30^\circ+\arcsin\left(\dfrac{2 \sqrt{39}}{13}\right)$

Next, use Law of Sines again:

$\dfrac{\sin \left(\angle ECF\right)}{x} = \dfrac{\sin\left(\angle AFC \right)}{6}$
$\dfrac{\sqrt{13}}{13x} = \dfrac{\dfrac{1}{2}\dfrac{\sqrt{13}}{13}+\dfrac{ \sqrt{3} }{2}\dfrac{2\sqrt{39}}{13}}{6}$
$\dfrac{\cancel{\sqrt{13}}}{\cancel{13}x} = \dfrac{7\cancel{\sqrt{13}}}{2\cdot 6\cdot \cancel{13}}$

$x = \dfrac{12}{7}$

5. ## Re: Equilateral last

Many Thanks for your help. But I must solve it with the classic method. I can't use trigonometry like you did. For example I have never used arcsin function in a problem unless I was asked to do so.

6. ## Re: Equilateral last

I found the solution without trigonometry.