Geometry - length of triangle question [![enter image description here][1]][1]
Could anyone please help me with this question
I can't seem to solve for the ?.
The answer provided for ? is 1.
Thanks!
When you see concurrent lines, Ceva's Theorem and Menelaus's Theorem are often the intended way.

Consider the grey triangle, applying Ceva's Theorem yields $$\frac{3}{x}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$
Again, consider the grey triangle with the black transversal line, applying Menelaus's Theorem yields $$\frac{5+x}{2}\cdot\frac{a}{b}\cdot\frac{d}{c}=1$$
Equating both equations gives $x=1$.

PS: I'm not sure why using other lines and triangles (for instance, the line between the line segments with lengths $3$ and $x$) doesn't work. If anyone would like to expand on this answer, please do. Thanks in advance!
Let us guess $\Delta ABC$  is an equilateral triangle with side length of 6.  F and D are the mid points of AC and BC respectively, CD=DB=AF=FC=3. Find a point E between D and B with  DE=1.
If we can prove  EF,AD and CH  intercept at the same point G, then   DE=1 is the solution.
Set some coordinates:
A(0,0),B(6,0),C(3,3$\sqrt3$), D(4.5,1.5$\sqrt3$), E(5,$\sqrt3$),F(1.5,1.5$\sqrt3$)
AE: y=$\frac{\sqrt3}{5}x$ (I)
AD:$\frac{\sqrt3}{3}x$  (II)
FB:  $-\sqrt3/3x$+2$\sqrt3$ (III)
EF:    $-\sqrt3/7x$+12$\sqrt3/7$ (IV)
Solve (I) and (III) to get the intercept point of AE and FB at H($\frac{15}{4},\frac{3\sqrt3}{4}$), then
HC:   $-3\sqrt3x+12\sqrt3$  (V)
Solve (II) and (V) to get the intercept point of AD and HC  at G$_1$(3.6,$\frac{6\sqrt3}{5}$)
Solve (II) and (IV) to get the intercept point of  AD  and EF intercept at G$_2$(3.6,$\frac{6\sqrt3}{5}$)
We  proved AD,EF and HC intercept at the same point  $G_1=G_2=G$.
Now  one can conclude the answer is indeed  DE=1.