\input zb-basic
\input zb-matheduc
\iteman{ZMATH 2015a.00610}
\itemau{Hung, Tran Quang}
\itemti{Two tangent circles from jigsawing quadrangle.}
\itemso{Forum Geom. 14, 247-248 (2014).}
\itemab
Let $ABC$ be an acute angled triangle. Van Lamoen has given the following construction: $P$ and $Q$ are a pair of isotomic points lying on $BC$, the perpendicular to $BC$ through $P$ intersects $AB$ at $P'$, the perpendicular to $BC$ through $Q$ intersects $AC$ at $Q'$, if we rotate $BPP'$ and $CQQ'$ about $P'$ and $Q'$ so that $P$ and $Q$ overlap, then also $B$ and $C$ overlap at a point $A'$. The quadrangle $AP'A'Q'$ is cyclic. If $S$ is the circumcenter of $AP'A'Q'$ and $T$ is the intersection of the tangents at $B$ and $C$ to the circumcircle of $ABC$ this paper proves that the circumcircle of $AP'A'Q'$ is tangent at $A'$ to the circle of center $T$ and radius $TB=TC$.
\itemrv{Antonio M. Oller (Zaragoza)}
\itemcc{G45}
\itemut{tangent circles; jigsawing a quadrangle}
\itemli{http://forumgeom.fau.edu/FG2014volume14/FG201425index.html}
\end