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The points A, B, C and D, in this particular order, lie on a circle. The chords AC and BD intersect in the point P, the line through $C$ perpendicular to AC and the line through $D$ perpendicular to BD intersect in the point Q.

How do you prove that the line AB and PQ are perpendicular to one another?

NOTE: the chords do not have to be perpendicular to one another!

Note

The following image makes both lines evidently nontangent to the circle. OP's last drawing makes the line perpendicular to BD seem tangent. Thanks GeoGebra :).

enter image description here

John
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  • Please add an image. "the perpendicular line" perpendicular to what and passing through what? – MickG Feb 24 '16 at 19:33
  • okay, done! @MickG – John Feb 24 '16 at 20:22
  • The drawing you added makes people think the chords must be perpendicular. Is that required? If not, better note it in the question (or include another drawing without this "inconveniency"). – MickG Feb 24 '16 at 20:35
  • you're right, sorry for the inconvenience! I noted it in the question – John Feb 24 '16 at 20:44
  • Other inconvenience: seems $P$ is the center of the circle, thus seems the chords are diameters, thus the lines $QD$ and $QC$, being orthogonal to diameters, are tangent to the circle, which need not be, right? – MickG Feb 24 '16 at 20:52
  • indeed! I just uploaded a different picture – John Feb 24 '16 at 21:27
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    Altogether, when making a diagram of a problem, it is best to avoid anything that looks like anything it's not guaranteed to be. That is, don't draw two lines at anywhere near a right angle unless they are guaranteed to be perpendicular; don't draw chords anywhere near the center of the circle unless they're supposed to be diameters; don't draw two chords nearly the same length unless they're really equal; and so forth. Your second attempt fixed one such ambiguity but left several others. – David K Feb 24 '16 at 21:28
  • alright, thanks guys, I'll try to fix it! I'm still in middle school, so I definitely got a lot to learn – John Feb 24 '16 at 21:40
  • The current version of the image is wrong $-$ the line perpendicular to $AC$ must go through $C$, not $A$. – TonyK Feb 26 '16 at 13:38

1 Answers1

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Hint: $PCQD$ is a cyclic quadrilateral. Can you figure out the solution from the figure?

enter image description here

Quang Hoang
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  • oh okay thanks! But how do I mathematically prove the perpendicularity from this conclusion? – John Feb 24 '16 at 23:12
  • Let $\gamma = \angle DQP = \angle DCP = \angle DCA = \angle DBA$. The lines $DP$ and $DQ$ are perpendicular. If you rotate them by an angle $\gamma$, in the same direction, you get the directions of lines $QP$ and $BA$. So these two lines are perpendicular. – David Feb 25 '16 at 18:59
  • wouldn't you have to rotate them by an angle alpha? – John Feb 26 '16 at 17:20
  • Why rotation? As shown in the picture $\alpha+\beta=90^\circ$. Now what happens in $\triangle BPM$? – Quang Hoang Feb 27 '16 at 00:44
  • alpha and beta are found in the same triangle? – John Mar 01 '16 at 15:43