$ABC$ is acuteangled triangle. Variable point $X$ lies on segment $AC$, and variable point $Y$ lies on the ray $BC$ but not segment $BC$, such that $\angle ABX+\angle CXY =90$. $T$ is projection of $B$ on the $XY$. Prove that all points $T$ lies on the line.
Problem
Source: St Petersburg Olympiad 2018, Grade 9, P3
Tags: geometry
PROF65
23.06.2018 02:58
Let the perpendicular to $BA$ cuts $AC$ at $B'$ then $XY$ is tangent to the circumcircle of $BB'X$ so $\angle BXT$ is constant hence $BXT$ remains similar which means that $X\mapsto T$ is a similarity therefore since $X$ remains in the line $AC$, $T$ also remains on the line , image of $AC$ by the similarity
RagvaloD
23.06.2018 09:10
PROF65 wrote: Let the perpendicular to $BA$ cuts $AC$ at $B'$ then $XY$ is tangent to the circumcircle of $BB'X$ so $\angle BXT$ is constant hence $BXT$ remains similar which means that $X\mapsto T$ is a similarity therefore since $X$ remains in the line $AC$, $T$ also remains on the line , image of $AC$ by the similarity Perpendicular to $BA$ at what point? Can you give image? Thanks
PROF65
23.06.2018 13:45
RagvaloD wrote: PROF65 wrote: Let the perpendicular to $BA$ cuts $AC$ at $B'$ then $XY$ is tangent to the circumcircle of $BB'X$ so $\angle BXT$ is constant hence $BXT$ remains similar which means that $X\mapsto T$ is a similarity therefore since $X$ remains in the line $AC$, $T$ also remains on the line , image of $AC$ by the similarity Perpendicular to $BA$ at what point? Can you give image? Thanks at $B$
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