Shanghai Math Competition Problem

Let $AF=a$ . We note that $\triangle AEF$ and $\triangle CEB$ are similar. Then $\dfrac {BC}{AF} = \dfrac {BE}{EF} = \dfrac 52 \implies BC = 2.5a$ . Also $\triangle GFD$ and $\triangle GBC$ are similar. Then

$\begin{aligned} \frac {FG}{BG} = \frac {FD}{BC} \\ \frac {FG}{BF+FG} & = \frac {BC-AF}{BC} \\ \frac {FG}{7+FG} & = \frac {2.5a-a}{2.5 a} = \frac 35 \\ 5FG & = 21 + 3FG \\ \implies FG & = \frac {21}2 = \boxed{10.5} \end{aligned}$

x := FG.

FG / EG = FD / BC

• FG / EG = x / (x+7)

• FD / BC = (AD - AF) / BC = 1 - 2/5 = 3/5

So, x / (x+7) = 3/5

5x = 3x + 21

x = 10.5

link to solution