Use the fact that the regular $3-$gon and the regular $5-$gon are constructible to show that the regular $15-$gon is constructible.
What is the best way to prove this? I have found a theorem that states that if $gcd(m,n)=1$ where $n-$gon and $m-$gon constructible, then the $mn-$ is also constructible.
Is there a better way to prove the question above?
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$\begingroup$Sure, if you can construct a regular 3-gon and a regular 5-gon, that means you can construct angles of 120° and 72°, which means you can construct the angle of $2\cdot 72° - 120°$, which is 24°, and that is all it takes to construct a regular 15-gon.
$\endgroup$ 1 $\begingroup$Given a fix circle $\Gamma$, you start construct a $5$-agon (regular) $ABCDE$ and the $3$-agon (regular) $FGH$ such that both not have common vertice. Now you consider the arc $FG$ that have measure $120^o$. With a compass you take the arc $AB$ that have measura $72^o$. With this position of compass you put one point of compass in the point $G$ and so draw the new arc that intersect $\Gamma$ in the point $I$. Note that the measure of arc $FI$ is the difference of the measures of the arcs $FG$ and $IG$ that are $120^o$ and $72^o$, respectivelly. Thus, the measure of the arc $FI$ is $120^o-72^o=48^o$. Now you draw the perpendiculat bisector $(r)$ of segment $FI$. This straight $(r)$ intersect $\Gamma$ in the point $J$. Note that $FJ$ is the side of $15$-agon (regular) inscribed in $\Gamma$.
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