Triangle geometry - synthetic proof I'm looking for a nice synthetic proof of the following fact.
Consider a non-isosceles triangle, pick a vertex. Assume that the median and the altitude passing through this vertex are isogonal conjugates (i.e. symmetric w.r.t. the bisector of the angle). Prove that the triangle is right-angled in this vertex.
 A: Let $\triangle ABC$ have the desired property at $A$. Let $M$, $P$, $N$ be the points where the median, angle bisector, and altitude from $A$ meet side $\overline{BC}$.
Since 
$$\angle BAP \cong \angle CAP \qquad \text{and} \qquad \angle MAP \cong \angle NAP$$
we have
$$\angle BAM \cong \angle CAN = C^\prime \qquad \text{and} \qquad
\angle CAM \cong \angle BAN = B^\prime$$
where $B^\prime$ and $C^\prime$ are respective complements of $B$ and $C$.
Invoking the Law of Sines in $\triangle BAM$ and $\triangle CAM$,
$$\underbrace{|\overline{AM}|\;\frac{\sin C^\prime}{\sin B} = |\overline{BM}|}_{\triangle BAM} = \frac{1}{2} |\overline{BC}| = \underbrace{|\overline{CM}| = |\overline{AM}|\; \frac{\sin B^\prime}{\sin C}}_{\triangle CAM}$$
Therefore,
$$\begin{align}
\sin B\sin B^\prime = \sin C \sin C^\prime \quad&\implies\quad \sin B\cos B = \sin C \cos C \\[4pt]
&\implies\quad \sin 2 B = \sin 2 C \\[4pt]
&\implies\quad B = C \quad\text{or}\quad 2B = \pi - 2C \\[4pt]
&\implies\quad B = C \quad\text{or}\quad B + C = \pi/2 \\[4pt]
&\implies\quad B = C \quad\text{or}\quad A = \pi/2
\end{align}$$
A: *

*First prove it conversely: that the condition holds for a right triangle.

 If the other two angles are $\alpha\le \beta=90^\circ-\alpha$, then one leg with the altitude makes angle $\alpha$ and the other leg with the median also makes angle $\alpha$, because of an isosceles triangle.


*Now if we have an arbitrary triangle $ABC$ with the given property at $A$, find the intersection $A'$ of the Thales circle raised over segment $BC$ with the half-line of side $BA$. Now if $A\ne A'$, the angle of side $AB$ with the altitude remains the same in triangles $ABC$ and $A'BC$ but the other angle in question will differ.
