Questions
In the triangle $ABC$ below we know the values of $\alpha$, $\beta$ and $\vert BD\vert$ in addition to knowing that AD is the median in the triangle, i.e. $\vert BD\vert = \vert DC\vert$.
Is it possible to analytically determine the other side lengths, i.e. $\vert AB\vert$, $\vert AD \vert$ and $\vert AC \vert$? If no, why not? Is the solution ambiguous?
My Approach
Since none of the triangles have enough information on their own, I have tried to utilize that the triangles $ABD$ and $ADC$ have the same areas to give me some insigts about the relationship between them:
\begin{align*}
\frac{1}{2} \vert AB \vert \vert AD\vert \sin(\alpha) = \frac{1}{2} \vert AD \vert \vert AC\vert \sin(\beta)
\end{align*}
simplifying to
\begin{align} \vert AB \vert \sin(\alpha) =\vert AC\vert \sin(\beta) \qquad(1), \end{align}
or equivalently
$$\frac{\sin(\alpha)}{\sin(\beta)} = \frac{\vert AB\vert}{\vert AC\vert}\qquad (2).$$ This is convenient, since I can compute the ratio on the left hand side.
To gain more information about $\vert AB\vert$ and $\vert AC\vert$, I derive and expression for each of them using the sinus theorem on the triangles ABD and ABC, resulting in the two equations:
\begin{align}
\vert BD \vert &= \frac{\sin(\alpha)}{\sin(\delta)}\vert AB\vert \\
\vert BD \vert &= \frac{\sin(\beta)}{\sin(\zeta)}\vert AC\vert
\end{align}
or in combination
\begin{align}
\frac{\vert AB\vert}{\vert AC \vert} = \frac{\sin(\beta)\sin(\delta)}{\sin(\alpha)\sin(\zeta)},
\end{align}
which can be substituted into Eq. (2) resulting in the relation
\begin{align}
\left(\frac{\sin(\alpha)}{\sin(\beta)} \right)^2=\frac{\sin(\delta)}{\sin(\zeta)},
\end{align}
or, when substituting $\delta = \pi – \zeta$
\begin{align}
\left(\frac{\sin(\alpha)}{\sin(\beta)} \right)^2=\frac{\sin(\pi – \zeta)}{\sin(\zeta)}.\qquad (3)
\end{align}
So when I go down this path my question becomes: Can I determine $\zeta$ analytically from Eq. (3)?
Best Answer
(Answer based on @Intelligentipauca's comments)
Apply the sine theorem to both triangle ABD and ADC we get the two equations
\begin{align} \frac{\vert AD\vert}{\vert BD\vert} &= \frac{\sin(\gamma)}{\sin(\alpha)} \\ \frac{\vert AD\vert}{\vert DC\vert} &= \frac{\sin(\epsilon)}{\sin(\beta)}. \end{align} Since $\vert BD\vert = \vert DC\vert$ we can combine the two equations to get
\begin{align} \frac{\sin(\beta)}{\sin(\alpha)} = \frac{\sin(\epsilon)}{\sin(\gamma)}. \end{align}
Utilizing the identity $\sin(x)=\sin(\pi-x)$ and that $\epsilon = \pi - \alpha - \beta - \gamma$ we get
\begin{align} \frac{\sin(\beta)}{\sin(\alpha)} = \frac{\sin(\alpha+\beta+\gamma)}{\sin(\gamma)}, \end{align} which has the solution \begin{align} \tan(\gamma) &= \frac{\sin(\alpha)\sin(\alpha+\beta)}{\sin(\beta)-\sin(\alpha)\cos(\alpha+\beta)}. \end{align} Knowing $\gamma$ the remaining values can be calculated using simple trigonometry.