2016 Eighth Romanian Master of Mathematics

Let $\sigma$ denote reflection in the line $BC$. Since $\angle BDF=\angle BAC=\angle CDE$, by concyclicity, the lines $DE$ and $DF$ are images of one another under $\sigma$, so the lines $AC$ and $DF$ meet at $P^{\prime }=\sigma (P)$, and the lines $AB$ and $DE$ meet at $Q^{\prime }=\sigma (Q)$. Consequently, the lines $PQ$ and $P^{\prime }{Q}^{\prime }=\sigma (PQ)$ meet at some (possibly ideal) point $R$ on the line $BC$. Since the pairs of lines $(CA,QD)$, $(AB,DP)$, $(BC,PQ)$ meet at three collinear points, namely $P^{\prime }$, $Q^{\prime }$, $R$ respectively, the triangles $ABC$ and $DPQ$ are perspective, i.e., the lines $AD,BP,CQ$ are concurrent, by the Desargues theorem.



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Alternative solution.

As in the first solution, $\sigma$ denotes reflection in the line $BC$, the lines $DE$ and $DF$ are images of one another under $\sigma$, the lines $AC$ and $DF$ meet at $P^{\prime }=\sigma (P)$, and the lines $AB$ and $DE$ meet at $Q^{\prime }=\sigma (Q)$. Let the line $AD$ meet the circle $ABC$ again at $M$. Letting $M^{\prime }=\sigma (M)$, it is sufficient to prove that the lines $D{M}^{\prime }=\sigma (AD)$, $B{P}^{\prime }=\sigma (BP)$ and $C{Q}^{\prime }=\sigma (CQ)$ are concurrent. Begin by noticing that $\angle (B{M}^{\prime },M^{\prime }D)=-\angle (BM,MA)=-\angle (BC,CA)=\angle (BF,FD)$, to infer that $M^{\prime }$ lies on the circle $BDF$. Similarly, $M^{\prime }$ lies on the circle $CDE$, so the line $D{M}^{\prime }$ is the radical axis of the circles $BDF$ and $CDE$. Since $P^{\prime }$ lies on the lines $AC$ and $DF$, it is the radical centre of the circles $ABC$, $ADC$, and $BDF$; hence the line $B{P}^{\prime }$ is the radical axis of the circles $BDF$ and $ABC$. Similarly, the line $C{Q}^{\prime }$ is the radical axis of the circles $CDE$ and $ABC$. So the conclusion follows: the lines $D{M}^{\prime }$, $B{P}^{\prime }$ and $C{Q}^{\prime }$ are concurrent at the radical centre of the circles $ABC$, $BDF$ and $CDE$; thus the lines $DM$, $B{P}^{\prime }$ and $C{Q}^{\prime }$ are also concurrent.

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Alternative solution.

As in the previous solutions, $\sigma$ denotes reflection in the line $BC$. Let the lines $BE$ and $CF$ meet at $X$. Due to the circles BDEA and CDFA, we have $\angle XBD=\angle EAD=\angle XFD$, so the quadrilateral BFXD is cyclic; similarly, the quadrilateral CEXD is cyclic. Hence $\angle XDB=\angle CFA=\angle CDA$, the lines DX and DA are therefore images of one another under $\sigma$, and $X^{\prime }=\sigma (X)$ lies on the line AD. Let $E^{\prime }=\sigma (E)$ and $F^{\prime }=\sigma (F)$, and apply the Pappus theorem to the hexagon $BP{F}^{\prime }CQ{E}^{\prime }$ to infer that $X^{\prime }$, D, and $BP\cap CQ$ are collinear. The conclusion follows.