Problem
There are two fixed points $A$ and $B$, and a moving point $P$ outside the line $AB$. Let $Q$ be the midpoint of $AP$ and $R$ be the midpoint of $BQ$. Then, $PR$ always passes through a fixed point on $AB$.
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Solution
Extend $PR$ and let the point where it intersects with $AB$ be $M$, and let the point where a line passing through $Q$ and parallel to $AB$ intersects with $PR$ be $C$.
Since $△PAM$ and $△PQC$ share $∠APM \ (=∠QPC)$, and $AM∥QC$,
$$∠PAM=∠PQC \qquad and \qquad ∠PMA=∠PCQ,$$
$$∴ \ △PAM \sim △PQC.$$
Since $PA∶PQ=2∶1$,
$$AM∶QC=2∶1. \qquad [1]$$
For $△RQC$ and $△RBM$,
$$∠QRC=∠BRM \qquad and \qquad RQ=RB.$$
Also, because $QC∥BM$,
$$∠RQC=∠RBM,$$
$$∴ \ △RQC≡△RBM,$$
$$∴ \ QC∶BM=1∶1. \qquad [2]$$
From $[1]$ and $[2]$,
$$AM∶BM=2∶1.$$
That is, $M$ is the point that divides the line segment $AB$ into $2∶1$, regardless of the location of $P$, and $PR$ always passes through the fixed point $M$.
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.28.