}

It is

\[
E A+A F+F D+D C=63+50+38+49=200=E C
\]

This implies that the points \(E, A, F, D, C\) lie on the same straight line in this order. \(A, B, F\) are the vertices of a right-angled triangle with the right angle \(\angle A B F\). This follows from

\[
A B^{2}+B F^{2}=30^{2}+40^{2}=50^{2}=A F^{2}
\]

by the converse of the Pythagorean theorem.

Let \(Q\) be the foot of the perpendicular from \(B\) to \(A F\). Since \(\triangle B F Q \sim \triangle A F B\), we have

\[
B Q=\frac{A B \cdot B F}{A F}=24 \quad \text { and } \quad Q F=\frac{B F^{2}}{A F}=32
\]

Thus, \(Q D=Q F+F D=70\). Finally, from the right-angled triangle \(\triangle B Q D\), by the Pythagorean theorem, we get

\[
B D^{2}=B Q^{2}+Q D^{2}=24^{2}+70^{2}=4\left(12^{2}+35^{2}\right)=4 \cdot 37^{2}=74^{2}
\]

and thus \(B D=74\). The distance from \(B\) to \(D\) is therefore \(74 \mathrm{~km}\).
\section*{Problem 3 - 061033}

From six locations \(A, B, C, D, E\) and \(F\), the following distances between them (in km) are known:

\[
A B=30, A E=63, A F=50, B F=40, C D=49, C E=200, D F=38
\]

What is the distance between \(B\) and \(D\)?