Geometry with perpendicularity

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Geometry with perpendicularity

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Source: 2018 Belarusian National Olympiad 9.3

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53 posts

#1 h Mar 29, 2019, 3:35 PM • 1 Y

Y by Adventure10

The bisector of angle $CAB$ of triangle $ABC$ intersects the side $CB$ at $L$ . The point $D$ is the foot of the perpendicular from $C$ to $AL$ and the point $E$ is the foot of perpendicular from $L$ to $AB$ . The lines $CB$ and $DE$ meet at $F$ .
Prove that $AF$ is an altitude of triangle $ABC$ .

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2125 posts

#3 h Mar 29, 2019, 4:29 PM • 2 Y

Y by Delta0001, Adventure10

Nice Problem

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.635, xmax = 12.765, ymin = -6.234, ymax = 8.266; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-5.34,5.09)--(-8.62,-3.13)--(3.32,-3.29)--cycle, linewidth(1.2) + rvwvcq); /* draw figures */ draw((-5.34,5.09)--(-8.62,-3.13), linewidth(1.2) + rvwvcq); draw((-8.62,-3.13)--(3.32,-3.29), linewidth(1.2) + rvwvcq); draw((3.32,-3.29)--(-5.34,5.09), linewidth(1.2) + rvwvcq); draw((-7.948887752602455,-1.448127233656149)--(-3.243067609749702,-4.696285290287367), linewidth(0.4)); draw((-5.34,5.09)--(-1.008579779428556,-4.217495865287583), linewidth(0.4)); draw((-5.450719858165978,-3.1724694156359665)--(-5.34,5.09), linewidth(0.4)); draw(circle((-4.452081309566209,0.9461250426742568), 4.237935731316973), linewidth(0.4) + linetype("4 4") + dtsfsf); draw(circle((-1.01,0.9), 6.025363059600641), linewidth(0.4) + linetype("4 4") + dtsfsf); draw((-3.243067609749702,-4.696285290287367)--(-5.34,5.09), linewidth(0.4)); draw((-3.243067609749702,-4.696285290287367)--(3.32,-3.29), linewidth(0.4)); draw((-1.008579779428556,-4.217495865287583)--(-7.948887752602455,-1.448127233656149), linewidth(0.4)); draw(circle((-3.174289889714279,0.43625206735620753), 5.13299815920288), linewidth(0.8)); draw(circle((-0.05330358953078493,1.8886624003178278), 6.180430516040682), linewidth(0.8)); /* dots and labels */ dot((-5.34,5.09),dotstyle); label("$A$", (-5.955,5.686), NE * labelscalefactor); dot((-8.62,-3.13),dotstyle); label("$B$", (-9.415,-3.834), NE * labelscalefactor); dot((3.32,-3.29),dotstyle); label("$C$", (3.885,-4.034), NE * labelscalefactor); dot((-3.243067609749702,-4.696285290287367),dotstyle); label("$D$", (-3.875,-5.534), NE * labelscalefactor); dot((-3.5641626191324196,-3.197749914651492),dotstyle); label("$L$", (-3.215,-2.634), NE * labelscalefactor); dot((-7.948887752602455,-1.448127233656149),dotstyle); label("$E$", (-9.055,-1.234), NE * labelscalefactor); dot((-5.450719858165978,-3.1724694156359665),dotstyle); label("$F$", (-6.115,-3.734), NE * labelscalefactor); dot((-1.008579779428556,-4.217495865287583),dotstyle); label("$K$", (-0.815,-4.934), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $EL \cap DC=K$ , then, obviously $EAKD$ is cyclic and $\angle ACB=\angle ALE=\angle DLK$ shows $ALKC$ cyclic $\implies$ $A$ is the miquel point of $EFDKCL$ and hence, $AFDC$ , $AEFL$ are also cyclic, $\implies$ $\angle ADC=\angle AFC=90^{\circ}$

This post has been edited 2 times. Last edited by AlastorMoody, Mar 29, 2019, 4:32 PM

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#4 h Mar 30, 2019, 12:10 PM • 2 Y

Y by Adventure10, Mango247

Let $F'$ be the feet of the perpendicular from $A$ to $BC$ .
Now, $AF'DC$ and $AEF'L$ are cyclic. Now, $\angle DF'C+\angle EF'C=\angle DAC+(180-\angle EAL)=\angle A/2+180-\angle A/2=180$
So, $E$ , $F'$ , $D$ are collinear so, $F'=F$ .

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162 posts

#5 h Apr 22, 2019, 10:57 AM • 2 Y

Y by Adventure10, Mango247

Let $K$ be the intersection of $AB$ and $CD$ . Triangle $ACK$ is isosceles. By symmetry, $\angle AKL=\angle ACB$
$EKDL$ is cyclic (right angles at $E$ and $D$ ), so $\angle EDL = \angle EKL = \angle ACB$
So $AFDC$ is cyclic, so $\angle AFC = \angle ADC = 90$

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MrReq

#6 h Sep 26, 2021, 9:38 AM

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just aply the well known lemma about intersection of the circles in spiral similarity $(\bigtriangleup AEL\sim \bigtriangleup ADC)$

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