In triangle $ABC$, $AD$ and $AH$ are the angle bisector and the altitude of vertex $A$, respectively. The perpendicular bisector of $AD$, intersects the semicircles with diameters $AB$ and $AC$ which are drawn outside triangle $ABC$ in $X$ and $Y$, respectively. Prove that the quadrilateral $XYDH$ is concyclic. Proposed by Mahan Malihi
Problem
Source: Iran TST 2013:TST 3,Day 2,Problem 2
Tags: Asymptote, geometry, circumcircle, angle bisector, perpendicular bisector, geometry proposed
War-Hammer
26.07.2013 21:48
Posted before : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=534218
AnonymousBunny
17.07.2014 20:57
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -4.300000000000004, xmax = 11.16000000000001, ymin = -2.860000000000003, ymax = 6.300000000000007; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((1.480000000000002,4.580000000000005)--(-1.240000000000001,-0.2000000000000002)--(2.760000000000003,0.02000000000000002)--cycle, zzttqq); /* draw figures */ draw((1.480000000000002,4.580000000000005)--(-1.240000000000001,-0.2000000000000002), zzttqq); draw((-1.240000000000001,-0.2000000000000002)--(2.760000000000003,0.02000000000000002), zzttqq); draw((2.760000000000003,0.02000000000000002)--(1.480000000000002,4.580000000000005), zzttqq); draw((1.480000000000002,4.580000000000005)--(1.733903940579748,-0.03643528326811396)); draw((1.480000000000002,4.580000000000005)--(0.9091733885298866,-0.08179546363085635)); draw(circle((0.1200000000000001,2.190000000000002), 2.749854541607613)); draw(circle((2.120000000000002,2.300000000000002), 2.368121618498512)); draw((-2.579991774191400,2.711291108018154)--(4.445180766286215,1.851074166378590)); /* dots and labels */ dot((1.480000000000002,4.580000000000005),dotstyle); label("$A$", (1.560000000000002,4.700000000000006), NE * labelscalefactor); dot((-1.240000000000001,-0.2000000000000002),dotstyle); label("$B$", (-1.160000000000001,-0.08000000000000008), NE * labelscalefactor); dot((2.760000000000003,0.02000000000000002),dotstyle); label("$C$", (2.840000000000003,0.1400000000000002), NE * labelscalefactor); dot((0.9091733885298866,-0.08179546363085635),dotstyle); label("$D$", (0.9800000000000011,0.04000000000000004), NW * labelscalefactor); dot((1.733903940579748,-0.03643528326811396),dotstyle); label("$H$", (1.820000000000002,0.08000000000000008), NW * labelscalefactor); dot((-2.579991774191400,2.711291108018154),dotstyle); label("$X$", (-2.500000000000003,2.840000000000003), NE * labelscalefactor); dot((4.445180766286215,1.851074166378590),dotstyle); label("$Y$", (4.520000000000006,1.980000000000002), NE * labelscalefactor); dot((0.2214379042003225,2.368262199293216),dotstyle); label("$Q$", (0.3000000000000003,2.480000000000003), NE * labelscalefactor); dot((2.167735484329565,2.129942337075933),dotstyle); label("$P$", (2.240000000000003,2.240000000000003), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the perpendicular bisector of $AD$ meet the circumcircle of $\triangle XDH$ at $Y'.$ Let $XY'$ meet $CA$ and $AB$ at points $P,Q$ respectively. Since $\angle AHB + \angle AXB = 90^{\circ} + 90^{\circ} = 180^{\circ},$ points $A,X,B,H$ are concyclic. Also, $\angle AXY' = \angle DXY$ from isoceles $\triangle XAD.$ We have \[\angle Y'AP = \angle PAY' + \angle AY'P \implies \angle PAY' = \angle Y'AP - \angle AY'P \\ \implies \angle PAY' = \angle DAY' - \angle DAC - \angle AY'P = \angle Y'DA - \angle DAC - \angle AY'P \\ = \angle AQP - \angle XAM = \angle XAM + \angle AXY' - \angle XAM \\ = \angle AXY' = \angle ADH = \angle Y'HB,\] so points $A,H,C,Y'$ are concyclic. Since $AH \perp BC,$ $H$ lies on the circle with diameter $AC,$ so $Y' = Y.$ $\blacksquare$
AleksaS
10.08.2020 21:00
Beautiful problem and I give beautiful but a bit long solution. Firstly, using angle bisector theorem we obtain that $$ \frac{BD}{CD} = \frac{AB}{AC} (1)$$Now let $M, N, P$ be midpoints of $AB$, $AC$ and $AD$, respectively. These points are collinear and they form a line that is parallel to line $BC$. In $\triangle ABD$, $MP$ is midline and so it follows that $MP = \frac{BD}{2}$. We also obtain in the same way that $NP = \frac{CD} {2}$. From this we get $$\frac{BD}{CD} = \frac{MP}{NP} (2) $$Because $AB$ and $AC$ are diameters it follows that $AB = 2MX$ and $AC = 2NY$ which implies that $$\frac{AB}{AC} = \frac{MX}{NY} (3)$$From $(1), (2), (3)$ we obtain that $$\frac{MP}{MX} = \frac{NP}{NY} (4)$$Easily we get that $\angle MPX = \angle NPY$. Now because of law of sines we get that $sin\angle PXM = sin\angle PYN$ which implies that $\angle PXM$ and $\angle PYN$ are suplement or equal. It is obvious that they are equal because $\angle PXM < \angle BAX = 90^\circ$ and $\angle PYN < \angle AYC = 90^\circ$. Now triangles $MPX$ and $PNY$ are similar and from there we write $\angle PMX = \angle XMB + \angle BMP = 180^\circ - \angle AMX + 180^\circ - \angle AMP =180^\circ - 2\angle ABX + 180^\circ - \angle ABC$ $(5)$ $$\angle PNY = \angle PNA + \angle ANY = \angle ACB + 2\angle ACY (6)$$From $(5) = (6)$ and some rewriting we get that $\angle ABX + \angle ACY = (180^\circ - \angle ABC - \angle BCA) + (90^\circ - \angle ABX) + (90^\circ - \angle ACY) = \angle BAC + \angle XAB + \angle CAY = \angle XAY$ $(7)$ It is obvious that $A, H, B, X$ and $A, Y, C, H$ are concyclic, and from there we get $\angle AHX = \angle ABX$ and $\angle AHY = \angle ACY$ which implies that $$\angle XAY = \angle AHX + \angle AHY = \angle XHY (8)$$Because line that contains $X, P, Y$ is perpendicular bisector of $AD$ it gives us that $$\angle XAY = \angle XDY (9)$$Now from $(8), (9)$ we get that $X, Y, D, H$ are concyclic as desired.
GaU1Er
21.03.2021 23:12
Let $K= AD \cap XY $ , and $O$ and $O'$ the midpoints of $AB$ and $AC$ respectively , clearly $K$ lies on $OO'$ , and since $O$ and $O'$ are the circumcenters of $(AXB)$ and $(AYC)$ respectively thus by the angle bisector theorem $\frac{KO}{KO'}=\frac{AO}{AO'}=\frac{OX}{O'Y}$ $\implies$ $O'Y\parallel{OX}$ . Since $\angle{AXB}=\angle{AHB}=\frac{\pi}{2}$ , hence $AXBH$ is cyclic , similary $AYCH$ is cyclic too , thus $\angle{XHY}=\angle{XHA}+\angle{AHY}=\angle{OBX}+\angle{O'CY}=\pi-\frac{\angle{XOB}+\angle{YO'C}}{2}=\pi-\frac{(\angle{XOK}-\angle{BOO'})+(\pi-\angle{AO'Y})}{2}=\pi-\frac{\angle{XOK}-(\pi-\angle{B})+\pi-(\angle{KO'Y}-\angle{C})}{2}=\pi-\frac{\angle{B}+\angle{C}}{2}=\frac{\angle{A}+\pi}{2}$. On the other hand , $\angle{XDY}=\angle{XAY}=\angle{A}+\angle{XHB}+\angle{YHC}=\angle{A}+\pi-\angle{XHY}=\angle{A}+\pi-\frac{\angle{A}+\pi}{2}=\angle{XHY}$.
guptaamitu1
30.05.2022 10:20
$\sqrt{bc}$ inversion just finishes it. $D^*$ would become the intersection of external angle bisector of $\angle X^*H^*Y^*$ with perpendicular bisector of segment $X^*Y^*$, and hence points $X^*,Y^*,D^*,H^*$ would be concyclic. Remark: This also shows that we only require $$ \frac{XA}{XD} = \frac{YA}{YD} $$for points $X,Y,D,H$ to be concylic (original problem is a special case when both sides equal $1$).
Number1048576
15.11.2022 15:06
Let $O_B$ be the semicircle with diameter $AB$, and let $O_C$ be the semicircle with diameter $AC$. Then let the external angle bisector of
$\angle BAC$ meet $O_B$ and $O_C$ at $F$ and $G$ respectively. Note that since $\angle BFA = 90$, $BF \parallel AD$, and similarly
$CG \parallel AD$. Now, we claim that $FC$, $GB$, $XY$ and $AD$ meet at a point.
Let $W = FC \cap GB$. By the angle bisector theorem, $\frac{AB}{AC} = \frac{BD}{DC}$. However, since $\triangle ABF \sim \triangle ACG$, so
$\frac{BD}{DC} = \frac{AB}{AC} = \frac{FB}{CG} = \frac{FA}{AG}$. Then, since $FB \parallel CG$, $\frac{FB}{CG} = \frac{BW}{GW}$. Therefore,
$\frac{BW}{GW} = \frac{BD}{DC}$, so $DW \parallel CG$, so $AD$ passes through $W$. However, note that
$\frac{BW}{BG} = \frac{DW}{GC}$, $\frac{AW}{GC} = \frac{FW}{FC}$. However, $\frac{FW}{FC} = \frac{BW}{BG}$, so $AW = DW$. This implies that
$W$ is the midpoint of $BC$, so $XY$ passes through $W$ as required.
Let $XY$ intersect $FB$ and $CG$ at $U$ and $V$ respectively; then $\frac{UF}{BU} = \frac{CV}{VG}$. Note that
$\angle BXF = \angle FAB = \angle CAG = \angle GYC$. However, note that there is only one point $X'$ with $\angle BX'F = \angle CAG$ such that
the perpendicular $U'$ from $X'$ to $BF$ satisfies $\frac{U'F}{BU'} = \frac{CV}{VG}$. However, there exists a point $X''$ satisfying this property such that $\triangle FBX'' \sim \triangle GYC$, so since we must have that $X'' = X$, $\triangle FBX \sim \triangle GYC$.
Then, letting $a = \angle BAC$, $\angle XFB = \angle XAB$ and $\angle FBX = \angle CGY = \angle CAY$, so $\angle XDY = \angle XAY = 90 + \frac{a}{2}$. On the other hand, $\angle AHX = \angle ABX = 90 - \angle XAB$, so $\angle XHY = 180 - \angle XAB - \angle CAY = 90 + \frac{a}{2}$.
Therefore, $\angle XDY = \angle XHY$, so $XDHY$ is cyclic as required.
jrsbr
21.02.2023 16:27
Let $M,N$ and $P$ be the midpoints of $AB$, $AC$ and $AD$, respectively. Since $M,N$ and $P$ are collinear, see that $$\frac{XM}{YN}=\frac{AM}{AN}=\frac{PM}{PN},$$and thus $\triangle XMP\sim\triangle YNP\implies XM\parallel YN\implies\angle ANY=\angle XMB+\angle A=180^{\circ}-\angle XMA+\angle A$ $$\angle ANY+\angle AMX=180^{\circ}+\angle A\implies\angle XHY=90^{\circ}+\frac{\angle A}{2}.$$On the other hand, we can easily chase some angles and prove $\angle XHY+\angle XDY=180^{\circ}+\angle A$, which concludes the problem. $\blacksquare$