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Problem Posts |  |
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QUESTION
(Canadian Mathematical Olympiad 2013, Problem 3) Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^{\circ}$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.
(Canadian Mathematical Olympiad 2013, Problem 3) Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^{\circ}$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.
Laconic Solution Sketch.
Using similar triangles, we can show that this common point is the foot of the altitude on $AB$ from $C$.
