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_e85ef69091c57cb633f3g-129.jpg?height=295&width=455&top_left_y=1544&top_left_x=1301)
By Pythagoras,

$$
B C^{2}=120^{2}+160^{2}=40000
$$

so $B C=200 \mathrm{~m}$.

Let $F B=x$. Then $C F=200-x$. From the given information,

$$
A C+C F=A B+B F
$$

$$
\begin{aligned}
120+200-x & =160+x \\
160 & =2 x \\
x & =80
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-232.jpg?height=353&width=542&top_left_y=306&top_left_x=1298)

Thus, the distance from $F$ to $B$ is $80 \mathrm{~m}$.

ANSWER: (D)

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