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$3$ median of a triangle $ABC$ divided this triangle into $6$ parts.

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How do I prove that circumcentre of each triangle is circumscribed?

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    I am sorry, but I am having a difficult time understanding your problem as posed. Can it be cleaned up?2012-12-27
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    @Haruboy15: Nice picture! I suggest you change the second sentence to "Prove that the circumcentre of each triangle is in the circumscibed circle of $\triangle ABC$."2012-12-27
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    Join the circumcenters in an appropriate order ---say $KJHIMN(K)$--- to get a hexagon whose opposite edges are parallel. (Eg, $KJ \parallel IM$ because both are $\perp$ to $AD$.) Each pair of opposite edges, therefore, intersects at a point on the "line at infinity"; by a converse of a special case of Pascal's Theorem (http://en.wikipedia.org/wiki/Pascal%27s_theorem), the points *at least* lie on a conic. The article "Hexagons with opposite sides parallel" ( http://www.jstor.org/stable/3621413 ) relates such hexagons to triangle Cevians and discusses general conditions that force circularity.2012-12-27

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