Consider a parallelogram $WXYZ$, with points $A$ and $B$ on sides $WX$ and $XY$ respectively, so that $\angle WAZ = \angle YBZ$. Let the midpoint of $WY$ be $M$. Prove that $OM$, where $O$ is the centre of the circle $AXB$, is perpendicular to $WY$.

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EDIT: In response to Mick's solution. I think you need to explain why the equal angles means the two lines are parallel. I think your solution breaks down when you say that KMN is a straight line without proof. Here is a picture where your first two paragraphs are correct, but doesn't solve the problem because the original angles are not the same.

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Let there be a red circle passing through $\triangle WXY$ with its circum-center at K. This circle will cut the blue circle (the circum-circle that passes through $\triangle AXB$ with center O) at G.

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Since M is the midpoint of the chord WY, $KM \bot WY$.

Since KO (extended) is the line of centers and GX is the common chord of the red and blue circles, we get $KO \bot GX$ at N and N is the midpoint of GX.

From the above - (1) KMN is a straight line joining the centers to the midpoints of two parallel chords; and (2) KON is the line of centers and therefore is also a straight line, we can say that KMON is straight line. Result follows because $\angle KMY$ is right-angled.


1) The green dotted line is the angle bisector of $\angle WZY$ in order to have $\angle WAZ = \angle YBZ$.

2) The dotted circle can be disregarded.

3) $\angle WAZ$ [edit: $= \angle YBZ$] restricts how the blue circle can be drawn but has no other bearings on the conclusion. [edit: The same color shaded angles will of course vary accordingly but they also have no bearings on the conclusion.]

  • $\begingroup$ Your third remark is a bit ambiguous. What do you mean it restricts the way in which it can be drawn? Surely it has an impact on whether O lies on that perpendicular? $\endgroup$ – Plato Jan 1 '17 at 19:35
  • $\begingroup$ In fact, having drawn the diagram myself, lines WY and XG are not parallel unless the two given angles are the same. $\endgroup$ – Plato Jan 1 '17 at 19:42
  • $\begingroup$ @Fermat Yes, the two given angle must be the same before we have WY // XG. By 'restrict', I mean the size of the circle AXB can vary (when the size of the $\angle WAZ [= \angle YBZ]$ varies) . I will edit the 3rd remark. $\endgroup$ – Mick Jan 2 '17 at 2:32
  • $\begingroup$ But have you proven that the two are parallel given that those two angles are the same? For example using an angle chase? $\endgroup$ – Plato Jan 2 '17 at 8:45
  • $\begingroup$ @Fermat I have spent quite a bit of time in angle chasing (in finding if $\angle GXW = \angle XWY$, for example), but not very successful. After that, I switched to proving KMN and KON are in fact the same straight line. After finding out that $\angle KMY = \angle KNX = 90^0$, I can then claim the two said lines are parallel. In the process, because I found that I need not use any one of the shaded angles, I therefore made remark #3. $\endgroup$ – Mick Jan 2 '17 at 9:55

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