Henry Ernest Dudeney/Modern Puzzles/138 - The Four Draughtsmen/Solution

by : $138$

 * The Four Draughtsmen

Solution

 * Dudeney-Modern-Puzzles-138-solution.png

Construct $AD$.

Construct $CE$ perpendicular to $AD$ and make $CE = AD$.

Because $AD$ stretches from one side of the board to its opposite, joining the centres of two squares, then the same applies to $CE$.

So $E$ is the centre of a square on the same row as $B$.

So join $EB$ and produce it in both directions.

Then:
 * construct $FG$ through $C$ parallel to $EB$
 * construct $FH$ through $A$ perpendicular to $EB$ and $FG$
 * construct $DG$ through $D$ parallel to $FH$.

As $H$ is the centre of a square on a board, we can mark $HE$ all round the square $HFG$ and discover the board is $10 \times 10$.

It remains to construct the actual board itself.