A circle with center $O$ is tangent to the sides of the angle with the vertex $A$ at the points B and C. Let M be a point on the larger of the two arcs $BC$ of this circle (different from $B$ and $C$) such that $M$ does not lie on the line $AO$. Lines $BM$ and $CM$ intersect the line $AO$ at the points $P$ and $Q$ respectively. Let $K$ be the foot of the perpendicular drawn from $P$ to $AC$ and $L$ be the foot of the perpendicular drawn from $Q$ to $AB$. Prove that the lines $OM$ and $KL$ are perpendicular.
Problem
Source: MOP 2005 Homework - Black Group #15
Tags: symmetry, geometry, perpendicular bisector
ThirdTimeLucky
28.05.2014 14:03
Suppose that the tangent at $ M $ to $ (O) $ meets $ AB,AC $ at $ X,Y $ respectively. Let $ AQ \cap XY=T $. Note that the pencil of lines $ (MX,MB,MA,MC) $ is harmonic and so we have, $ (AQTP)=-1 $. Note that $ \frac{AL}{AQ}=\frac{AK}{AP} \implies $ $ \frac{AL}{AK}=\frac{AQ}{AP}=\frac{QT}{TP} $. To show that $ OM \perp KL $, it suffices to show that $ KL \parallel XY $, or $ \frac{AL}{AK} = \frac{AX}{AY}=\frac{TX}{TY} $ $ \iff \frac{TQ}{TP} = \frac{TX}{TY} $. Hence, it only suffices to show that $ XQ \parallel YP $. But this is true as both $ XQ,YP $ are perpendicular to $ AO $.
Particle
07.06.2014 14:22
For brevity, denote $\angle BMC=m, \angle MBC=b, \angle MCB=c$. WLOG assume $b>c$. Let $D,E$ denote the feet of perpendiculars from $B$ to $MC$ and $C$ to $MB$, respectively. Suppose $DE\cap AB=L'$ and $Q'L'\perp AB$ with $Q'\in MC$. Then notice that $BL'DQ'$ concyclic (with diameter BQ'). Also $\angle DL'B=\angle (DE,BC)+\angle CBA=b-c+m$. So $\angle BQ'D=180-b+c-m=2c$. So $\angle Q'BC=c$ Hence $Q'$ lies on the perpendicular bisector of $BC$, implying $Q=Q'$ and $L=L'$. By symmetry $K\in DE$, too. It is very well known that $DE\perp OM$. So $KL\perp OM$.
ethical_man
08.12.2016 02:32
Particle wrote: $\angle BQ'D=180-b+c-m=2c$. Isn't $\angle BQ'D = \angle DL'B$? because $BDQ'L'$ is cyclic? So shouldn't we have $\angle BQ'D = b - c + m$ instead of $\angle BQ'D = 180 - b + c - m$?