If a circumference cuts a triangle $ABC$ at its sides $BC$, $CA$ and $AB$ at points $P, P'; Q, Q'; R, R'$; respectively (so twice on each side, and if $AP, BQ$ and $CR$ are concurrent (intersect at a single point), prove that $AP'$, $BQ'$ and $CR'$ are concurrent.not concurrent in this picture, but to illustrate first parte


By Ceva, $AP,BQ$ and $CR$ are concurrent if and only if $\frac{|BP|}{|CP|}\cdot\frac{|CQ|}{|AQ|}\cdot\frac{|AR|}{|BR|}=1$.

Similarly $AP',BQ'$ and $CR'$ are concurrent iff $\frac{|BP'|}{|CP'|}\cdot\frac{|CQ'|}{|AQ'|}\cdot\frac{|AR'|}{|BR'|}=1$.

So it suffices to show that $\frac{|BP|}{|CP|}\cdot\frac{|CQ|}{|AQ|}\cdot\frac{|AR|}{|BR|}=\frac{|CP'|}{|BP'|}\cdot\frac{|AQ'|}{|CQ'|}\cdot\frac{|BR'|}{|AR'|}$.

This follows from the equalities

$|CP|\cdot|CP'|=|CQ|\cdot |CQ'|$, $|AQ|\cdot|AQ'|=|AR|\cdot |AR'|$ and $|BP|\cdot|BP'|=|BR|\cdot |BR'|$.

  • $\begingroup$ It's worth noting that "the equalities" come from the secant-secant aspect of the "Power of a Point" theorem. $\endgroup$ – Blue Jun 12 '16 at 21:59

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