Consider equilateral triangle ABC and points D and E such that $BD = \frac{BC} 3$, DE belongs to (BC), and E is the midpoint of (AD). If DF is perepndicular on AD, F belongs to AC, prove that EF = AD.
my ideas
MY DRAWING
Okey so all i could thought of is the formula of the median. I thought we can apply it in the triangle AFD with the median FE. Another thing i thought of it that $BD = \frac{BC} 3$, that knida reminds me of the propriety of the G center.
Note: sorry for the lack of illustrations, feel free to add one or give suggestions on how to add them.
My proof uses the Pythagorean theorem a lot, not the prettiest proof ever :)
So, if the sides of $\triangle ABC$ are of length $2x$, we can find $AD=\sqrt{(\sqrt3\cdot x)^2+\left(\frac13x\right)^2}=\frac23\sqrt7\cdot x$ using the Pythagorean theorem. Also note the area of $\triangle ABC$ is $\sqrt3\cdot x^2$.
(If you wonder where $\sqrt3\cdot x$ and $\frac13x$ came from, draw $P$ as the midpoint of $BC$ and we can see that $\triangle ADP$ is a right triangle)
Now, I created a point $G$ in $AD$ so that $CG\parallel FD$.
As the area of $\triangle ACD$ is $\frac23\sqrt3\cdot x^2$, we can find that $CG=\frac{\frac43\sqrt3\cdot x^2}{AD}=\frac27\sqrt{21}\cdot x$.
Again, using the Pythagorean theorem, we can find that $AG=\sqrt{(2x)^2-\left(\frac27\sqrt{21}\cdot x\right)^2}=\frac47\sqrt7\cdot x$.
Notice that $\triangle ACG\sim\triangle AFD$, which means $\frac{CG}{FD}=\frac{AG}{AD}$. So:
$\frac{\frac27\sqrt{21}\cdot x}{FD}=\frac{\frac47\sqrt7\cdot x}{\frac23\sqrt7\cdot x}$
We can find that $FD=\frac13\sqrt{21}\cdot x$.
As we know $ED=\frac12AD=\frac13\sqrt7\cdot x$, using the Pythagorean theorem
(I promise this is the last time)we can find that $EF=\sqrt{\left(\frac13\sqrt{21}\cdot x\right)^2+\left(\frac13\sqrt7\cdot x\right)^2}=\frac23\sqrt7\cdot x$...which is the same as $AD!$ Definitely not the most elegant proof (considering all the Pythagorases), but that's what I came up with.