AoPS - Problem

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Tags: geometry, projections, bisects segment, Kharkiv, altitudes

parmenides51

22.03.2020 15:50

In the acute-angled triangle $ABC$ , let $CD, AE$ be the altitudes. Points $F$ and $G$ are the projections of $A$ and $C$ on the line $DE$ , respectively, $H$ and $K$ are the projections of $D$ and $E$ on the line $AC$ , respectively. The lines $HF$ and $KG$ intersect at point $P$ . Prove that line $BP$ bisects the segment $DE$ .

BestChoice123

22.03.2020 17:04

Could you check your problem again? I think something has gone wrong...

parmenides51

22.03.2020 17:11

the altitudes are $CD, AE$ and not $CD,BE$ , thanks

BestChoice123

23.03.2020 05:19

: In

$\Delta PQR$ , let

$QS,RT$ be the altitudes. Let

$O$ be the midpoint of

$QR$ . Points

$U,V,W$ are the projections of

$Q,R,O$ onto line

$TS$ then

$TU=SV$ [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 = -5.384971727358799, xmax = 8.693045096271753, ymin = -3.443880600546527, ymax = 4.566025523243271; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-2.079501504569394,0.2540107405762819)--(-1.7999110417696564,0.37048203594517803)--(-1.9163823371385527,0.6500724987449156)--(-2.19597279993829,0.5336012033760195)--cycle, linewidth(0.7)); draw((2.237961882107378,2.3806817898437895)--(2.3544331774762743,2.101091327044052)--(2.6340236402760118,2.2175626224129483)--(2.5175523449071155,2.4971530852126858)--cycle, linewidth(0.7)); draw((0.895877339797679,1.6941234937973888)--(1.1118236746109111,1.4817474828775783)--(1.3241996855307219,1.6976938176908103)--(1.1082533507174896,1.910069828610621)--cycle, linewidth(0.7)); draw((-0.9036252933613929,0.8412945190735308)--(-0.6242353524568391,0.7243430314608028)--(-0.5072838648441113,1.0037329723653565)--(-0.786673805748665,1.1206844599780845)--cycle, linewidth(0.7)); draw((-0.11880069031532489,1.3989058489254562)--(-0.0023293949464285313,1.1193153861257186)--(0.27726106785330873,1.2357866814946148)--(0.1607897724844126,1.5153771442943524)--cycle, linewidth(0.7)); /* draw figures */ draw((0,3)--(-1.6156041396930112,-0.8595793377908167), linewidth(0.7) + wrwrwr); draw((-1.6156041396930112,-0.8595793377908167)--(3.9099460740443357,-0.8453014302617797), linewidth(0.7) + wrwrwr); draw((3.9099460740443357,-0.8453014302617797)--(0,3), linewidth(0.7) + wrwrwr); draw((-1.6156041396930112,-0.8595793377908167)--(-2.19597279993829,0.5336012033760195), linewidth(0.7) + wrwrwr); draw((-2.19597279993829,0.5336012033760195)--(2.5175523449071155,2.4971530852126858), linewidth(0.7) + wrwrwr); draw((2.5175523449071155,2.4971530852126858)--(3.9099460740443357,-0.8453014302617797), linewidth(0.7) + wrwrwr); draw((-1.6156041396930112,-0.8595793377908167)--(1.1082533507174896,1.910069828610621), linewidth(0.7) + wrwrwr); draw((3.9099460740443357,-0.8453014302617797)--(-0.786673805748665,1.1206844599780845), linewidth(0.7) + wrwrwr); draw((1.1471709671756622,-0.8524403840262982)--(0.1607897724844126,1.5153771442943524), linewidth(0.7) + wrwrwr); draw(circle((1.1471709671756622,-0.8524403840262982), 2.7627843303077535), linewidth(0.7) + wrwrwr); /* dots and labels */ dot((0,3),dotstyle); label("

$P$ ", (0.05491104120432554,3.1382347703395634), NE * labelscalefactor); dot((-1.6156041396930112,-0.8595793377908167),dotstyle); label("

$Q$ ", (-1.8583285676866421,-1.0880258582554099), NE * labelscalefactor); dot((3.9099460740443357,-0.8453014302617797),dotstyle); label("

$R$ ", (3.9956135192185576,-0.8310235227327426), NE * labelscalefactor); dot((-0.786673805748665,1.1206844599780845),linewidth(4pt) + dotstyle); label("

$T$ ", (-0.9588203933573066,1.25355097650667), NE * labelscalefactor); dot((1.1082533507174896,1.910069828610621),linewidth(4pt) + dotstyle); label("

$S$ ", (1.1685878284692173,2.024557983074672), NE * labelscalefactor); dot((1.1471709671756622,-0.8524403840262982),linewidth(4pt) + dotstyle); label("

$O$ ", (1.2114215510563284,-0.7453560775585202), NE * labelscalefactor); dot((0.1607897724844126,1.5153771442943524),linewidth(4pt) + dotstyle); label("

$W$ ", (0.2833575616689187,1.3392184216808924), NE * labelscalefactor); dot((2.5175523449071155,2.4971530852126858),linewidth(4pt) + dotstyle); label("

$V$ ", (2.5678227663148503,2.609952191765192), NE * labelscalefactor); dot((-2.19597279993829,0.5336012033760195),linewidth(4pt) + dotstyle); label("

$U$ ", (-2.1438867182673835,0.653878860287113), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Proof Note that

$QUVR$ is a right trapezoid thus

$W$ is the midpoint of

$UV$ Since

$QTSR$ is a cyclic quadrilateral with circumcenter

$O$ , we have

$WT=WS\implies UT=SV$

[asy][asy]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 = -12.809202841741886, xmax = 7.921632970754376, ymin = -4.907023331008777, ymax = 4.987929048015788; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-3.479965676744615,-2.3763944993842596)--(-3.801895153666741,-2.378750080922714)--(-3.7995395721282863,-2.70067955784484)--(-3.47761009520616,-2.6983239763063858)--cycle, linewidth(0.7) + qqwuqq); draw((-4.554889101668133,0.4154596999240554)--(-4.413928019760133,0.7048972756861951)--(-4.703365595522273,0.8458583575941948)--(-4.8443266774302725,0.5564207818320551)--cycle, linewidth(0.7) + qqwuqq); draw((-2.4093558889701523,-4.410775638059076)--(-2.5339993635200293,-4.113945482704415)--(-2.8308295188746904,-4.238588957254292)--(-2.7061860443248134,-4.535419112608953)--cycle, linewidth(0.7) + qqwuqq); draw((-5.491107253761741,2.0966857627799618)--(-5.19427709840708,2.221329237329839)--(-5.318920572956957,2.5181593926845)--(-5.615750728311618,2.393515918134623)--cycle, linewidth(0.7) + qqwuqq); draw((-2.997506763900295,2.2003791013313183)--(-2.783452541449304,1.9599109605587)--(-2.5429844006766857,2.173965183009691)--(-2.7570386231276767,2.4144333237823092)--cycle, linewidth(0.7) + qqwuqq); draw((-0.7768884333991597,-0.29425966443030305)--(-0.5628342109481685,-0.5347278052029214)--(-0.3223660701755503,-0.3206735827519303)--(-0.5364202926265413,-0.080205441979312)--cycle, linewidth(0.7) + qqwuqq); /* draw figures */ draw((-3.521302755909007,3.273006319749262)--(-6.44,-2.72), linewidth(0.7)); draw((-6.44,-2.72)--(1.76,-2.66), linewidth(0.7)); draw((1.76,-2.66)--(-3.521302755909007,3.273006319749262), linewidth(0.7)); draw((-6.44,-2.72)--(0.6907982367713352,2.4396613983669346), linewidth(0.7)); draw((-5.615750728311618,2.393515918134623)--(-3.521302755909007,3.273006319749262), linewidth(0.7)); draw((-5.615750728311618,2.393515918134623)--(-2.7061860443248134,-4.535419112608953), linewidth(0.7)); draw((-2.7061860443248134,-4.535419112608953)--(1.76,-2.66), linewidth(0.7)); draw((-3.47761009520616,-2.6983239763063858)--(-0.5364202926265413,-0.080205441979312), linewidth(0.7)); draw((1.76,-2.66)--(-4.8443266774302725,0.5564207818320551), linewidth(0.7)); draw((-4.8443266774302725,0.5564207818320551)--(-2.7570386231276767,2.4144333237823092), linewidth(0.7)); draw((-3.47761009520616,-2.6983239763063858)--(-3.521302755909007,3.273006319749262), linewidth(0.7)); draw((-2.7061860443248134,-4.535419112608953)--(0.6907982367713352,2.4396613983669346), linewidth(0.7)); draw((0.6907982367713352,2.4396613983669346)--(-5.615750728311618,2.393515918134623), linewidth(0.7)); draw((-3.943671198167062,2.4057506464039724)--(-1.8055305650616023,-2.6860892480370357), linewidth(0.7)); draw(circle((-0.8806513779545034,0.30650315987463084), 3.9715463861623443), linewidth(0.7)); /* dots and labels */ dot((-3.521302755909007,3.273006319749262),dotstyle); label("

$A$ ", (-3.4605975265898374,3.4247693930471836), NE * labelscalefactor); dot((-6.44,-2.72),dotstyle); label("

$B$ ", (-6.723503602495179,-2.99480860745495), NE * labelscalefactor); dot((1.76,-2.66),dotstyle); label("

$C$ ", (1.8207574241778777,-2.509166772901597), NE * labelscalefactor); dot((-4.8443266774302725,0.5564207818320551),linewidth(4pt) + dotstyle); label("

$D$ ", (-5.160343947526574,0.3591553124291435), NE * labelscalefactor); dot((-3.47761009520616,-2.6983239763063858),linewidth(4pt) + dotstyle); label("

$E$ ", (-3.6882421365367217,-2.9796323001251577), NE * labelscalefactor); dot((-5.615750728311618,2.393515918134623),linewidth(4pt) + dotstyle); label("

$F$ ", (-5.554927938101173,2.514190953259647), NE * labelscalefactor); dot((-2.7061860443248134,-4.535419112608953),linewidth(4pt) + dotstyle); label("

$G$ ", (-2.6866058527704313,-4.8614944090194), NE * labelscalefactor); dot((-2.7570386231276767,2.4144333237823092),linewidth(4pt) + dotstyle); label("

$H$ ", (-2.7017821601002234,2.5293672605894395), NE * labelscalefactor); dot((-0.5364202926265413,-0.080205441979312),linewidth(4pt) + dotstyle); label("

$K$ ", (-0.28874929466325,-0.14166282945400166), NE * labelscalefactor); dot((0.6907982367713352,2.4396613983669346),linewidth(4pt) + dotstyle); label("

$P$ ", (0.7584159110924177,2.559719875249024), NE * labelscalefactor); dot((-4.160968386318216,-1.0709515972371657),linewidth(4pt) + dotstyle); label("

$I$ ", (-4.158707663760283,-0.8397729666244464), NE * labelscalefactor); dot((-3.943671198167062,2.4057506464039724),linewidth(4pt) + dotstyle); label("

$J$ ", (-4.00694459046236,2.013372811376502), NE * labelscalefactor); dot((-1.8055305650616023,-2.6860892480370357),linewidth(4pt) + dotstyle); label("

$L$ ", (-1.730498490993517,-2.873398148816612), NE * labelscalefactor); dot((-2.874600881614332,-0.14016930081653203),linewidth(4pt) + dotstyle); label("

$M$ ", (-2.899074155387523,0.05562916583329795), NE * labelscalefactor); dot((-0.8806513779545034,0.30650315987463084),linewidth(4pt) + dotstyle); label("

$N$ ", (-0.81992005120598,0.4350368490781049), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let

$J=HF\cap AD$ and

$L=CE\cap KG$ We have

$\angle AHJ=\angle FDJ=\angle BDE=\angle ECA\implies PJ\parallel BL$ Similarly, we also have

$BJ\parallel PL\implies BJPL$ is a parallelogram Construct

$M=JL\cap BP$ then

$M$ is the midpoint of

$JL$ This means we are done if we can show that

$DE\parallel JL$ Utilizing the Lemma, we get

$DF=EG\implies \Delta DFJ=\Delta ELG\implies FJ\parallel =EL\implies EFJL$ is a parallelogram thus

$DE\parallel JL$ Hence done