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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I suspect the end of the problem should have been "Prove the MN is perpendicular to QP". – Henning Makholm Jun 8 '12 at 13:58
It actually appears to be the perpendicular bisector of QP, which would explain the "symmetrical" part. – MJD Jun 8 '12 at 14:01
@Adam: What have you tried? – MJD Jun 8 '12 at 14:01
I have no idea where to start from. – Adam Jun 8 '12 at 17:41

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.

Claim 1. $PQXO$ lie on a circle.

Just check that $\angle OQX +\angle OPX = \pi/2+\pi/2=\pi.$

Claim 2. The center $U$ of this circle lies on $MN$.

$XO$ is a diameter of the cercle, so $U$ is the midpoint of $XO$.

Claim 3. $MN$ passes through the mid point of $PQ$

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.

Therefore, the remaining task is to show that the midpoints of $AB$, $CD$ and $XO$ are collinear. (Then the connection of the cycle center $M$ with the midpoint of a chord is automatically perpendicular to the chord.)

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 )

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There are two proofs of the theorem on cut-the-knot: cut-the-knot.org/Curriculum/Geometry/Quadri.shtml#explanation – Phira Jun 8 '12 at 22:07
Do you have any idea how can I prove it without using the Gauss line? – Adam Jun 9 '12 at 11:44
@AdamAndersson You can go for a calculatory proof. It is relatively easy to see $MN$ bisects $PQ$ using coordinate calculation, but for the orthogonality, I didn't see any approach that I wanted to calculate to the end. I would not expect a really simple proof because it is obviously easily equivalent to the existence of a Gauss line for a cyclic quadrilateral. – Phira Jun 9 '12 at 11:57
There must be a way to prove this using some supplementary circle. – Adam Jun 14 '12 at 18:09

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.

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