On the diagonal $AC$ of the rhombus $ABCD$, a point $E$ is taken, which is different from points $A$ and $C$, and on the lines $AB$ and $BC$ are points $N$ and $M$, respectively, with $AE = NE$ and $CE = ME$. Let $K$ be the intersection point of lines $AM$ and $CN$. Prove that points $K, E$ and $D$ are collinear.
Problem
Source: Oral Moscow Team MO 2017 IX p5 / All-Russian MO 1993 Regional (Round 4) 9.7
Tags: geometry, rhombus, collinear, equal segments
rafaello
19.09.2020 16:30
We see that $E$ is the spiral similarity centre of $AN$ and $MC$, thus $E$ lies on $ANK$ and $KMC$. Hence, $K$ is Miquel point wrt $ABC$, thus $BMKN$ is cyclic, which means that $AKCD$ is cyclic and since $BD$ bisects $AC$, then $KD$ is the angle bisector of $\angle AKC$. Also, \begin{align*} \measuredangle AKE &=\measuredangle ANE=\measuredangle EAN\\ &=\measuredangle MCE=\measuredangle EMC\\&=\measuredangle EKC. \end{align*}Thus, $KE$ is the angle bisector of $\angle AKC$. Hence, we are done. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11cm); 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 = -0.83, xmax = 5.11, ymin = -5.29, ymax = 4.83; /* image dimensions */ /* draw figures */ draw((0,0)--(4,0), linewidth(1)); draw((2,4)--(2,-4), linewidth(1)); draw((0,0)--(2,-4), linewidth(1)); draw((2,-4)--(4,0), linewidth(1)); draw((4,0)--(2,4), linewidth(1)); draw((2,4)--(0,0), linewidth(1)); draw((xmin, 0.6560424966799467*xmin + 0)--(xmax, 0.6560424966799467*xmax + 0), linewidth(1)); /* line */ draw((xmin, -0.3612750885478158*xmin + 1.4451003541912633)--(xmax, -0.3612750885478158*xmax + 1.4451003541912633), linewidth(1)); /* line */ draw((xmin, -8.51063829787234*xmin + 13.021276595744682)--(xmax, -8.51063829787234*xmax + 13.021276595744682), linewidth(1)); /* line */ draw(circle((2,-1.5), 2.5), linewidth(1) + linetype("2 2")); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label("$A$", (0.09,0.17), NE * labelscalefactor); dot((4,0),dotstyle); label("$C$", (4.09,0.21), NE * labelscalefactor); dot((2,4),dotstyle); label("$B$", (2.09,4.21), NE * labelscalefactor); dot((2,-4),dotstyle); label("$D$", (2.09,-3.79), NE * labelscalefactor); dot((1.53,0),dotstyle); label("$E$", (1.61,0.21), NE * labelscalefactor); dot((0.612,1.224),linewidth(4pt) + dotstyle); label("$N$", (0.69,1.39), NE * labelscalefactor); dot((3.012,1.976),linewidth(4pt) + dotstyle); label("$M$", (3.09,2.13), NE * labelscalefactor); dot((1.4205007120443378,0.9319088336652095),linewidth(4pt) + dotstyle); label("$K$", (1.51,1.09), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
FarrukhKhayitboyev
14.09.2024 17:54
Observe that $K$ is the Miquel Point of $\triangle ABC$, So $ANKE$ and $CMKE$ are cyclic quadrialterals. $EK$ will be radical axis of $w_1(ANKE)$ and $w_2(CMKE)$ We will prove that $DA$ is tangent to $w_1(ANKE)$, $\angle DAC= \angle BAC = \angle ANE$, symmetrically $DC$ is tangent to $w_2(CMKE)$. From that we conclude $D$ should lie on radical axis of these two circles, meaning $K,E,D$ are collinear and we are done.