$(BUL 4)$ Let $M$ be the point inside the right-angled triangle $ABC (\angle C = 90^{\circ})$ such that $\angle MAB = \angle MBC = \angle MCA =\phi.$ Let $\Psi$ be the acute angle between the medians of $AC$ and $BC.$ Prove that $\frac{\sin(\phi+\Psi)}{\sin(\phi-\Psi)}= 5.$
RHS should be -5 as LHS is defined. Rename $\angle MAB=\angle MBC=\angle MCA=\omega$ the Brocard angle of $\triangle ABC$ and let $G$ be the centroid of $\triangle ABC.$ Tangent of the Brocard angle can be found using the fact that a median and a symmedian issuing from two vertices of any triangle intersect on a Brocard ray issuing from the third vertex.
$\tan \omega=\frac{4[\triangle ABC]}{AB^2+BC^2+CA^2}=\frac{4[\triangle ABC]}{AB^2+AB^2}=\frac{2[\triangle ABC]}{AB^2}$
$\cot \Psi= - \frac{GA^2+GB^2-AB^2}{4[\triangle GAB]}= - \frac{3(\frac{2}{3}AB^2-GC^2-AB^2)}{4[\triangle ABC]}=\frac{AB^2}{3[\triangle ABC]}$
$\frac{ \sin ( \omega+ \Psi)}{\sin (\omega- \Psi)}=\frac{ \sin \omega \cos \Psi+\cos \omega \sin \Psi}{\sin \omega \cos \Psi-\cos \omega \sin \Psi}=\frac{\tan \omega+ \tan \Psi}{\tan \omega- \tan \Psi}=\frac{ 2+3}{ 2-3}=-5.$