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In the non-isosceles triangle $ABC$ an altitude from $A$ meets side $BC$ in $D$ . Let $M$ be the midpoint of $BC$ and let $N$ be the reflection of $M$ in $D$ . The circumcircle of triangle $AMN$ intersects the side $AB$ in $P \neq A$ and the side $AC$ in $Q \neq A$ . Prove that $AN,BQ$ and $CP$ are concurrent.

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We complete the reflection by reflecting everything across $AD$ .

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Now it's a simple matter of angle chasing to show that $PQCB$ must be cyclic. After that, I am lost. I have tried drawing the other altitudes, extended $PQ$ in the hopes of using harmonic bundles.They have not yielded particularly delightful results.

How can I proceed after this?

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  • $\begingroup$ Use the Power of Point $B$ (and $C$) with respect to the circle to get $|BN||BM| = |BP||BA|$ (and $|CM||CN|=|CQ||CA|$). Then Ceva's Theorem can get you there without too much trouble. $\endgroup$
    – Blue
    Mar 30, 2016 at 16:53
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    $\begingroup$ @Blue , wow, nice ! And there I was trying to use poles and polars;forgot to try out the really basic stuff. Thanks. Do you mind putting that as an answer for acceptance later on? $\endgroup$
    – rah4927
    Mar 30, 2016 at 17:42

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(Converting comment to answer, as requested.)

Use the Power of Point $B$ (and $C$) with respect to the circle to get $$|\overline{BN}||\overline{BM}| = |\overline{BP}||\overline{BA}| \qquad\text{and}\qquad |\overline{CM}||\overline{CN}| = |\overline{CQ}||\overline{CA}|$$ Then Ceva's Theorem can get you there without too much trouble.

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