In the diagram, $A B C$ represents a triangular jogging path. Jack jogs along the path from $A$ to $B$ to $F$. Jill jogs from $A$ to $C$ to $F$. Each jogs the same distance. The distance from $F$ to $B$, in metres, is
(A) 40
(B) 120
(C) 100
(D) 80
(E) 200

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-129.jpg?height=298&width=447&top_left_y=2019&top_left_x=1311)
By Pythagoras,

$$
\begin{aligned}
B C^{2} & =120^{2}+160^{2} \\
B C^{2} & =40000 \\
B C & =200
\end{aligned}
$$

Let $F B=x$. Then $F C=200-x$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-210.jpg?height=336&width=542&top_left_y=1859&top_left_x=1168)

So Jack jogs $A B+B F=160+x$, and Jill jogs $A C+C F=120+200-x$. Since they jog the same distance,

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-210.jpg?height=180&width=403&top_left_y=2295&top_left_x=365)

Therefore, it is 80 metres from $B$ to $F$.

ANSWER: (D)