The given angle conditions imply no configuration issues. Let medians $BM$ and $CN$ intersect at centroid $R$. Note that we wish to prove $RP\cdot RN = RQ \cdot RM$. Since $\angle CQB = \angle CPB=90^{\circ}$, $CBPQ$ is cyclic. PoP with respect to the circumcircle of $CBPQ$, yields $RP = \frac{RQ\cdot RB}{RC}$ and $RQ = \frac{RP\cdot RC}{RB}$. Thus we wish to prove $$\frac{RQ\cdot RB}{RC}\cdot RN = \frac{RP\cdot RC}{RB} \cdot RM \implies RB^2\cdot RQ \cdot RN = RC^2 \cdot RP \cdot RM.$$Note that $\triangle CRQ \sim \triangle BRP\implies \frac{CR}{RQ} = \frac{BR}{RP}$, thus we wish to prove $RB\cdot RN = RC \cdot RM$. This is true, since $RN = \frac{RC}{2}$ and $RM=\frac{RB}{2}$. $\blacksquare$

Thanks for the good problems, parmenides51 - I swear half of my geometry handouts would be barren of examples/exercises if it weren't for the problems you post.

I just want to note that the weird angle conditions do not matter - my solution (and GeoGebra) do not care about configurations.

all the problems that I post, are always mentioned their source (either a magazine or a contest) there are a few problems that I have posted without source where the aops' admins are to blame for losing the source as they moved them from HSO to HSM and the source field was lost (as HSM has no source field) I am glad that someone uses them, to provide a different problem set from the well known problems, as many Russian Constests are a lot worthy Edit: If there is any problem posted in aops by me sourceless, just send me the link (PM) to search in my page for the correct source