Finding area of triangles inside an isososeles triangle

Question

Hello,
I was going through some Olympiad math questions, when I came across this question

In $\triangle ABC, AB = AC, \angle A = 120°.$ Points $D, E, F$ are on segments $BC, CA, AB$ respectively such that $CD = CE, BD = BF, DE = 10, FD = 6$. Find area of $\triangle DEF$.

This question left me in a haze, for the fact that the diagram can't be drawn using scale-compass and it even can't be drawn on geogebra (The point E comes outside $\overline{AC}$, while the question states that it is on the line)
Using Extension of Pythagoras Theorem ($a^2 = b^2 + c^2 + bc \sqrt 3$) where the values of the number rooted is correspondent to the angles or cosine rule, I get irrational numbers as the lengths and am not able to proceed further. I intended using Herons formula due to the lack of side lengths but am not able to find the length of $\overline{FE}$.
Can I get some hints to solve it using plain geometry(I know it has some thing to do with excentres but I'm not able to make any observations on this point)
ANY HELP IS APPRECIATED

Answer 1

It's $$\frac{6\cdot10\sin30^{\circ}}{2}=15.$$

The geometric solution.

Let $FK$ be an altitude of $\Delta FDE$.

Thus, since $$\measuredangle FDE=180^{\circ}-2\cdot75^{\circ}=30^{\circ},$$ we obtain: $$FK=\frac{1}{2}FD=3$$ and $$S_{\Delta DEF}=\frac{3\cdot10}{2}=15.$$

Answer 2

Your discovery that the problem is inconsistent is correct. $BFD$ and $CDE$ are $30-75-75$ triangles, which gives $BD=\frac 3{\sin 15^\circ}\approx 11.59, CD=\frac 5{\sin 15^\circ}\approx 19.31$ and $E$ lies beyond $A$ from $C$. Report that fact and you are done.

If you let $E$ be on the extension of $CA$ and want to find the area of $DEF$ anyway, note that $\angle FDE=30^\circ$, use the law of cosines to get the length of $FE$ and you have all three sides for Heron's formula.