Let $M$ be the midpoint of side $BC$ of acute triangle $ABC$. The circle centered at $M$ passing through $A$ intersects the lines $AB$ and $AC$ again at $P$ and $Q$, respectively. The tangents to this circle at $P$ and $Q$ meet at $D$. Prove that the perpendicular bisector of $BC$ bisects segment $AD$.
Problem
Source: 2022 NZMO - New Zealand Maths Olympiad Round 1 p7
Tags: geometry, bisects segment
Rijul saini
29.11.2023 19:35
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.114cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.358181818181821, xmax = 20.02181818181819, ymin = -5.124545454545452, ymax = 8.77545454545454; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ttqqqq = rgb(0.2,0.,0.); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-5.828955988570143,6.573186280364034)--(-5.6802397280650085,6.175840745770565)--(-5.28289419347154,6.3245570062756995)--(-5.431610453976674,6.721902540869168)--cycle, linewidth(0.8) + rvwvcq); /* draw figures */ draw(circle((-3.96,2.79), 4.587373976470634), linewidth(0.8) + qqwuqq); draw((-3.7,7.37)--(-7.163220907953347,6.073805081738337), linewidth(0.8) + wvvxds); draw((-7.152595897895664,-0.5041359159481826)--(-0.21395703521420034,0.14214008943471304), linewidth(0.8) + rvwvcq); draw((-7.163220907953347,6.073805081738337)--(-3.305938957585367,-4.2322216663843895), linewidth(0.8) + dbwrru); draw((-3.7,7.37)--(-7.152595897895664,-0.5041359159481826), linewidth(0.8) + rvwvcq); draw((-6.051429059253376,2.0072316492189772)--(-1.8685709407466213,3.5727683507810233), linewidth(0.8) + wvvxds); draw((-3.7,7.37)--(-0.21395703521420034,0.14214008943471304), linewidth(0.8) + rvwvcq); draw((-0.21395703521420034,0.14214008943471304)--(-3.305938957585367,-4.2322216663843895), linewidth(0.8) + wrwrwr); draw((-3.305938957585367,-4.2322216663843895)--(-7.152595897895664,-0.5041359159481826), linewidth(0.8) + wrwrwr); draw((-3.7,7.37)--(-4.22,-1.79), linewidth(0.8) + ttqqqq); draw((-3.7,7.37)--(-3.305938957585367,-4.2322216663843895), linewidth(0.8) + dtsfsf); draw((-3.5029694787926835,1.5688891668078053)--(-5.431610453976674,6.721902540869168), linewidth(0.8) + dbwrru); /* dots and labels */ dot((-3.96,2.79),dotstyle); label("$M$", (-4.618181818181819,2.8554545454545437), NE * labelscalefactor); dot((-3.7,7.37),dotstyle); label("$A$", (-3.618181818181819,7.575454545454542), NE * labelscalefactor); dot((-7.163220907953347,6.073805081738337),dotstyle); label("$S$", (-7.498181818181821,6.3354545454545415), NE * labelscalefactor); dot((-6.051429059253376,2.0072316492189772),linewidth(4.pt) + dotstyle); label("$B$", (-6.65818181818182,1.935454545454544), NE * labelscalefactor); dot((-1.8685709407466213,3.5727683507810233),linewidth(4.pt) + dotstyle); label("$C$", (-1.7981818181818183,3.735454545454543), NE * labelscalefactor); dot((-7.152595897895664,-0.5041359159481826),linewidth(4.pt) + dotstyle); label("$P$", (-7.73818181818182,-0.8845454545454545), NE * labelscalefactor); dot((-0.21395703521420034,0.14214008943471304),linewidth(4.pt) + dotstyle); label("$Q$", (0.14181818181818248,-0.20454545454545486), NE * labelscalefactor); dot((-3.305938957585367,-4.2322216663843895),linewidth(4.pt) + dotstyle); label("$D$", (-3.0781818181818186,-4.544545454545453), NE * labelscalefactor); dot((-4.22,-1.79),linewidth(4.pt) + dotstyle); label("$R$", (-4.71818181818182,-2.364545454545454), NE * labelscalefactor); dot((-3.5029694787926835,1.5688891668078053),linewidth(4.pt) + dotstyle); label("$E$", (-3.1981818181818187,1.6754545454545442), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Call circle with centre $M$ and radius $MA$ as $\omega$. Take $R$ to be the point diametrically opposite $A$ in $\omega$, and $S$ to be the point on $\omega$ so that $AS$ is parallel to $BC$. Let $E$ be the midpoint of $AD$. $M$ is midpoint of $BC$ so $(B,C; M, \infty) = -1$ is harmonic, projecting from $A$ and taking intersection with $\omega$, we get $(P,Q; R,S)$ is harmonic, so $PRQS$ is a harmonic quadrilateral. Therefore the tangents at $P$ and $Q$ intersects on $RS$, and thus $D,R,S$ are collinear. Let the perpendicular bisector of $BC$ be $\ell$. $AS$ is parallel to $BC$ so $\ell \perp AS$. $\ell$ passes through $M$, therefore $\ell$ is also the perpendicular bisector of $AS$. Finally, it remains to observe that $\triangle ASD$ is a right triangle so $E$ is the circumcentre, and hence the perpendicular bisector of $AS$ passes through $E$. Hence, $\ell$, passes through $E$.