Construction of a Triangle from $a,\;$ $m_a,\;$ and $l_a$

Jack D'Aurizio
25 March, 2016

By Stewart's theorem we have:

$\displaystyle b^2+c^2 = \frac{4m_a^2+a^2}{2},\qquad l_a^2 = bc\left(1-\frac{a^2}{(b+c)^2}\right). $

Hence it follows that $\Pi=bc\;$ is a solution of the second-degree equation:

$ l_a^2 \left(4m_a^2+a^2+4\Pi\right) = \Pi\left(4m_a^2-a^2+4\Pi\right).$

So $\Pi\;$ can be found through straightedge and compass once $a,m_a,l_a\;$ are known.

Given $b^2+c^2\;$ and $\Pi\;$, since $(b\pm c)^2 = b^2+c^2\pm 2\Pi,\;$ $b\;$ and $c\;$ can be found by straightedge and compass, through the formula:

$\displaystyle \frac{1}{2}\left(\sqrt{\frac{4m_a^2+a^2}{2}+2\Pi}\pm \sqrt{\frac{4m_a^2+a^2}{2}-2\Pi}\right).$

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