In an obtuse triangle $ABC$ $(AB>AC)$,$O$ is the circumcentre and $D,E,F$ are the midpoints of $BC,CA,AB$ respectively.Median $AD$ intersects $OF$ and $OE$ at $M$ and $N$ respectively.$BM$ meets $CN$ at point $P$.Prove that $OP\perp AP$
Clearly $O$ is the incenter of $\triangle MNP\implies OP$ internal bisector of $\angle MPN$. As constructed, $BM=AM\ (\ 1\ ), CN=AN\ (\ 2\ )$.
Menelaos for $\triangle BCP$ and transversal $\overline{DMN}$ with $(1),(2)$ gives $AP$ external bisector of $\angle MPN$, and we are done!
Best regards,
sunken rock
Complex bashing works Athough the expression for $m$ is rather ugly, it turns out that $m-b$ is very nice for factoring and conjugating, and similarly $n-c$. Solving these two equations we magically get $p=\frac{bc-a^2}{b+c-2a}$.