Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.
Problem
Source: Sharygin First Round 2013, Problem 6
Tags: geometry, trapezoid, circumcircle, Asymptote, perpendicular bisector, geometry proposed
aZpElr68Cb51U51qy9OM
09.04.2013 02:10
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.000000000000000, xmax = 21.00000000000000, ymin = -8.000000000000000, ymax = 6.000000000000000; /* image dimensions */ /* draw figures */ draw((4.980000000000006,3.480000000000004)--(2.880000000000003,-6.500000000000007)); draw((2.880000000000003,-6.500000000000007)--(16.76000000000002,-6.500000000000007)); draw((8.579010154576137,3.480000000000004)--(16.76000000000002,-6.500000000000007)); draw((4.980000000000006,3.480000000000004)--(16.76000000000002,-6.500000000000007)); draw((8.579010154576137,3.480000000000004)--(2.880000000000003,-6.500000000000007)); draw(circle((3.001284355006989,-1.314578872296060), 5.186839323365234)); draw(circle((13.59822072228109,-0.7486960660346604), 6.563104840078217)); draw((1.022568710013975,3.480000000000004)--(18.61743128998604,3.480000000000004)); draw((9.820000000000010,3.480000000000004)--(16.76000000000002,-6.500000000000007)); draw((9.820000000000010,3.480000000000004)--(2.880000000000003,-6.500000000000007)); /* dots and labels */ dot((4.980000000000006,3.480000000000004),dotstyle); label("$A$", (5.060000000000006,3.600000000000004), NE * labelscalefactor); dot((2.880000000000003,-6.500000000000007),dotstyle); label("$B$", (2.380000000000003,-7.200000000000008), NE * labelscalefactor); dot((16.76000000000002,-6.500000000000007),dotstyle); label("$C$", (16.72000000000002,-7.260000000000008), NE * labelscalefactor); dot((8.579010154576137,3.480000000000004),dotstyle); label("$D$", (7.900000000000009,3.760000000000004), NE * labelscalefactor); dot((7.405557239567553,1.425071201113405),dotstyle); label("$P$", (6.520000000000008,1.000000000000000), NE * labelscalefactor); dot((1.022568710013975,3.480000000000004),dotstyle); label("$X$", (0.4000000000000005,3.660000000000004), NE * labelscalefactor); dot((18.61743128998604,3.480000000000004),dotstyle); label("$Y$", (18.70000000000002,3.600000000000004), NE * labelscalefactor); dot((9.820000000000010,3.480000000000004),dotstyle); label("$M$", (9.900000000000011,3.600000000000004), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $\angle DYC = \angle BPC = \angle AXB$ because quadrilaterals $APBX$ and $DYCP$ are cyclic. Since $XY \parallel BC$, quadrilateral $BCYX$ is an isosceles trapezoid. Thus if $M$ lies on the perpendicular bisector of $XY$, it also lies on the perpendicular bisector of $BC$, implying that $BM = CM$.