I need a hint on this problem. ABCD is inscribed quadrilateral. Diagonals AC and BD intersect at point O. OP and OQ are the perpendiculars from O to BC and AD. M and N are the midpoints of AB and CD. Prove that MN is perpendicular bisector of QP.
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The perpendicular bisector suggests that we should look for a nice circle passing through $P$ and $Q$.
The most symmetric third point to choose is $O$ and looking at the figure shows additional properties:
Let $X$ be the intersection of $AD$ and $BC$ and denote by $\angle A$ etc the angles of the original quadrilateral.
Just check that $\angle OQX +\angle OPX = \pi/2+\pi/2=\pi.$
$XO$ is a diameter of the cercle, so $U$ is the midpoint of $XO$.
The triangles $BOC$ and $AOD$ are similar and $P$ and $Q$ are the corresponding feet of the altitudes. Therefore, $P=xB+yC$ and $Q=xA+yD$ for some real numbers $x+y=1$.
Now, $(P+Q)/2=x(A+B)/2 +y (C+D)/2 =xM+yN$ clearly lies on the line $MN$ as claimed.
Since these six points are the points of a complete quadrilateral, this is the well-known theorem that the mid-points of the diagonals of a complete quadrilateral are collinear (see for example http://mathworld.wolfram.com/CompleteQuadrilateral.html )
I have not laboured all the details, but you might be able to prove it alternatively arguing as follows: $MN$ shall be a perpendicular bisector, if $\Delta QNP$ (or $\Delta QMP$) is isosceles AND $MN$ bisects $\angle QNP$ ($\angle QMP$). You will have proved this by showing that $\Delta QNM = \Delta PNM$. Since these triangles have one common side it is enough to prove that their other sides are proportionate to each other. This requires proving that a couple of other triangles in the picture are isometric. The fact that $\angle ADC + \angle ABC =\angle DAB +\angle DCB = \pi$ which easily follows from the fact that the quadrilateral is inscribed in a circle should provide some help.