Let $\triangle{ABC}$ a triangle with $\angle{CAB}=90^{\circ}$ and $L$ a point on the segment $BC$. The circumcircle of triangle $\triangle{ABL}$ intersects $AC$ at $M$ and the circumcircle of triangle $\triangle{CAL}$ intersects $AB$ at $N$. Show that $L$, $M$ and $N$ are collinear.
Problem
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Tags: geometry, circumcircle, geometry proposed
WakeUp
31.01.2011 19:00
This is problem 3 from this years British MO (Round 1), posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=381276
Pedantic Anarchy
19.02.2011 23:03
MLB=MAB=90=CAB=NLB, then MLB=NLB.
vanu1996
22.12.2013 14:05
easy to see that $M$ is the orthocenter of $CNB$.
jayme
22.12.2013 15:14
Dear Mathlinkers, 1. the two circles in question are orthogonal 2. By a converse theorem, we are done Sincerely Jean-Louis
jayme
11.11.2021 10:16
Dear Mathlinkers, here Problem 5, p. 14 Sincerely Jean-Louis