Two friends, A and B, start walking from a common point. A walks 20 km towards…

Concept: When two straight-line legs are walked at right angles to each other — first p km, then q km turned through 90° — the resultant straight-line distance from the start is the hypotenuse of that right triangle: √(p² + q²) (Pythagoras' theorem). In this problem, the accompanying diagram shows A's stated north-east leg as exactly this same diagonal — the hypotenuse of the very right triangle traced by B's two legs — so because B's Pythagorean resultant matches A's stated distance, the diagram shows the two routes finishing at the identical point.

Application:

1. B walks two legs at right angles to each other: 16 km east, then 12 km north.

2. By the Pythagorean theorem, B's straight-line distance from the starting point is √(16² + 12²) = √(256 + 144) = √400 = 20 km — and, per the diagram, this diagonal is exactly the north-east leg A is stated to walk.

3. A walks a single 20 km leg directly towards the north-east from the very same starting point.

4. Since the diagram identifies A's 20 km north-east leg with this very same diagonal — the hypotenuse of B's own 16 km + 12 km right triangle — both walkers' paths terminate at the identical point, even though they take different routes to get there.

Cross-check: The diagram confirms this directly: the hypotenuse of the right triangle formed by B's two legs is labelled as the finishing point of both A and B.

Result: So A and B end up at the same place; the distance between them is 0 km.