# Solution

(related to Problem: The Joiner's Problem)

Nothing could be easier than the solution to this puzzle — when you know how to do it. And yet it is apt to perplex the novice a good deal if he wants to do it in the fewest possible pieces — three. All you have to do is to find the point $A,$ midway between $B$ and $C,$ and then cut from $A$ to $D$ and from $A$ to $E.$ The three pieces then form a square in the manner shown. Of course, the proportions of the original figure must be correct; thus the triangle $BEF$ is just a quarter of the square $BCDF.$ Draw lines from $B$ to $D$ and from $C$ to $F$ and this will be clear.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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