Given two line segments A and B of arbitrary length, is it possible to construct a segment C such that the proportion of C to A is equal to the proportion of [the area of a square with side B] to [the area of a square with side A]?
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Hint If $DEF$ is a right triangle, $D=90^\circ$ and $DH$ is the altitude, then $$DH^2=EH \cdot FH$$ Now construct a right triangle such that $EH=1$ and $DH$ is the given distance.
N. S.
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Is $A$ a point, a segment or an angle? – Jam Oct 22 '16 at 14:47
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$A$ is a vertex of the triangle $ABC$. – N. S. Oct 22 '16 at 14:51
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1Right. I raised the question since you've given it an angle but OPs defined it as a segment. Could possibly be confusing for them. – Jam Oct 22 '16 at 14:53
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@Jam Rewrote it. – N. S. Oct 22 '16 at 15:18