In the quadrilateral $ABCD$, shown in fig. , the equations are true: $\angle ABC = \angle BCD$ and $2AB = CD$. On the side $BC$, a point $X$ is selected such that $\angle BAX = \angle CDA$. Prove that $AX = AD$.
Problem
Source:
Tags: geometry, equal angles, equal segments
rafaello
05.09.2020 12:21
Extend $CD$, $AB$, $BC$ and $AD$, and let $E=AB\cap CD$, $F=AD\cap BC$ and $G=AX\cap DC$. Since $\angle FCD = \angle ABX$ and $\angle CDF = \angle BAX$, we get that $CDF\sim BAX$, thus $$\frac{CD}{AB}=\frac{FD}{AX}=2$$Also since $\angle CDA = \angle BAX\implies \angle ADE = \angle GAE$ and $\angle DEA =\angle AEG$, we get that $ADE\sim GAE$, thus $\angle EAD =\angle EGA$, hence $$\angle CGX =\angle BAF$$and since $\angle FCD = \angle ABX\implies \angle GCX = \angle ABF$, we get that $GCX\sim ABF$, thus $\angle FXA = \angle CXG =\angle BFA=\angle XFA$, hence $AX=AF$ and because $FD=AF+AD=2AX$, we get that $AX=AD$ and we are done. [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 = -1.0753333333333321, xmax = 16.602190476190458, ymin = -9.915923809523807, ymax = 5.756457142857137; /* image dimensions */ pen zzttqq = rgb(255,0,0); draw((1.61,3.32)--(1.81,-1.54)--(4.37,-0.68)--(6.43,2.14)--cycle, linewidth(1) + zzttqq); /* draw figures */ draw((1.61,3.32)--(1.81,-1.54), linewidth(1) + zzttqq); draw((1.81,-1.54)--(4.37,-0.68), linewidth(1) + zzttqq); draw((4.37,-0.68)--(6.43,2.14), linewidth(1) + zzttqq); draw((6.43,2.14)--(1.61,3.32), linewidth(1) + zzttqq); draw((1.6660415222573501,1.958191009146393)--(4.37,-0.68), linewidth(1)); draw((1.9130220507583493,-4.043435833427891)--(1.81,-1.54), linewidth(1)); draw((4.37,-0.68)--(1.9130220507583493,-4.043435833427891), linewidth(1)); draw((4.37,-0.68)--(10.043826578182417,1.2260511161081555), linewidth(1)); draw((10.043826578182417,1.2260511161081555)--(6.43,2.14), linewidth(1)); draw((1.61,3.32)--(-0.17847287680615193,3.757841907599846), linewidth(1)); draw((-0.17847287680615193,3.757841907599846)--(1.6660415222573501,1.958191009146393), linewidth(1)); /* dots and labels */ dot((1.61,3.32),dotstyle); label("$C$", (1.8747619047619033,3.5899809523809476), NE * labelscalefactor); dot((1.81,-1.54),dotstyle); label("$B$", (1.2985714285714274,-2.3563047619047635), NE * labelscalefactor); dot((4.37,-0.68),dotstyle); label("$A$", (4.31780952380952,-1.6187809523809544), NE * labelscalefactor); dot((6.43,2.14),dotstyle); label("$D$", (6.622571428571423,1.3313142857142821), NE * labelscalefactor); dot((1.6660415222573501,1.958191009146393),dotstyle); label("$X$", (0.860666666666666,1.4465523809523773), NE * labelscalefactor); dot((1.9130220507583493,-4.043435833427891),linewidth(4pt) + dotstyle); label("$F$", (2.2204761904761887,-4.3614476190476195), NE * labelscalefactor); dot((10.043826578182417,1.2260511161081555),dotstyle); label("$E$", (10.125809523809513,0.38636190476190163), NE * labelscalefactor); dot((-0.17847287680615193,3.757841907599846),linewidth(4pt) + dotstyle); label("$G$", (0.03095238095238107,4.073980952380947), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]