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Let $A$, $B$, $C$, $D$, and $E$ be points on a circle. For any three points, we draw the line going through the centroid of the triangle formed by these three points that is perpendicular to the line passing through the other two points. (For example, we draw the line going through the centroid of triangle $BDE$ that is perpendicular to $AC$.) In this way, we draw a total of $\binom{5}{3} = 10$ lines. Show that all 10 lines pass through the same point.

I realize this question has been asked before, but the explanations did not make much sense to me. Is there a simpler method somewhere where an explanation can be easily understood by a precalculus student? In other words, I'm asking for an alternative proof to this: Prove that secants of a circle pass through a common point and this: Prove the lines are concurrent (using vectors)?

Thanks.

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  • $\begingroup$ Could you link to the original question and ask specifically for an alternative-proof? $\endgroup$ – AlexR Aug 16 '15 at 18:27
  • $\begingroup$ @AlexR: Edited. $\endgroup$ – Mathy Person Aug 16 '15 at 18:30
  • $\begingroup$ @MathyPerson I've added a picture to make it easier to see what you're asking. Feel free to remove it if it's not to your liking. The edit is pending review so the picture won't be up immediately. $\endgroup$ – wltrup Aug 16 '15 at 19:42

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