Triangle geometry - synthetic proof

Question

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.

Answer 1

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, $$\begin{align} \sin B\sin B^\prime = \sin C \sin C^\prime \quad&\implies\quad \sin B\cos B = \sin C \cos C \\[4pt] &\implies\quad \sin 2 B = \sin 2 C \\[4pt] &\implies\quad B = C \quad\text{or}\quad 2B = \pi - 2C \\[4pt] &\implies\quad B = C \quad\text{or}\quad B + C = \pi/2 \\[4pt] &\implies\quad B = C \quad\text{or}\quad A = \pi/2 \end{align}$$

Answer 2

1. 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.

2. 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.