PE$ bisects <BPC wanted, circumcircle, incenter related

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PE$ bisects <BPC wanted, circumcircle, incenter related

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Source: 2022 FOFO = Fake OMA Foros Open, Easter p6 https://artofproblemsolving.com/community/c2688147_

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#1 h Apr 21, 2022, 1:54 PM • 1 Y

Y by ImSh95

Let $\omega$ be the circumcircle of an acute triangle $ABC$ , $D$ be the midpoint of arc $BAC$ , and $I$ be the incenter of triangle $ABC$ . The line $DI$ intersects $BC$ at $E$ and $\omega$ for second time at $F$ . Let $P$ be at $AF$ such that $EP$ is parallel to $AI$ . Prove that $PE$ bisects $\angle BPC$ .

This post has been edited 1 time. Last edited by parmenides51, Apr 21, 2022, 1:55 PM

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#2 h Apr 21, 2022, 4:27 PM • 1 Y

Y by ImSh95

$[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 = -31.55065425182996, xmax = 21.39097618986839, ymin = -21.292482231054976, ymax = 10.361912993876768; /* image dimensions */ /* draw figures */ draw(circle((-4.344087469074979,-3.9670536982634936), 10.02959552855595), linewidth(0.4) + red); draw((-9.148785977635322,4.836789707219564)--(-14.36894368356669,-3.6587610691001062), linewidth(0.4)); draw((-14.36894368356669,-3.6587610691001062)--(4.635159860248812,-8.435375963416387), linewidth(0.4)); draw((4.635159860248812,-8.435375963416387)--(-9.148785977635322,4.836789707219564), linewidth(0.4)); draw((-9.148785977635322,4.836789707219564)--(-8.377666341117513,-1.21850999526025), linewidth(0.4)); draw((-9.148785977635322,4.836789707219564)--(-13.862854380121021,-7.127095305521932), linewidth(0.4)); draw((-1.8992281791900671,5.759993619350055)--(-13.862854380121021,-7.127095305521932), linewidth(0.4)); draw((-14.36894368356669,-3.6587610691001062)--(-11.701507044281756,-1.64178860421456), linewidth(0.4)); draw((-11.701507044281756,-1.64178860421456)--(4.635159860248812,-8.435375963416387), linewidth(0.4)); draw((-11.701507044281756,-1.64178860421456)--(-11.347957364959688,-4.418075404118023), linewidth(0.4)); draw((-14.36894368356669,-3.6587610691001062)--(-13.862854380121021,-7.127095305521932), linewidth(0.4)); draw((-13.862854380121021,-7.127095305521932)--(4.635159860248812,-8.435375963416387), linewidth(0.4)); draw((-1.8992281791900671,5.759993619350055)--(-4.866891911658938,-6.047068516258246), linewidth(0.4)); draw((-1.8992281791900671,5.759993619350055)--(-9.148785977635322,4.836789707219564), linewidth(0.4)); draw((-18.798087102726157,-2.5455113766722337)--(-14.36894368356669,-3.6587610691001062), linewidth(0.4)); draw((-18.798087102726157,-2.5455113766722337)--(-8.377666341117513,-1.21850999526025), linewidth(0.4)); draw(circle((-15.073022233842924,-3.4817933903951204), 3.840928570880885), linewidth(0.4) + linetype("4 4") + red); draw((-18.798087102726157,-2.5455113766722337)--(-13.862854380121021,-7.127095305521932), linewidth(0.4)); draw((-6.788946758959895,-13.694101015877054)--(-4.866891911658938,-6.047068516258246), linewidth(0.4)); draw((-8.377666341117513,-1.21850999526025)--(-6.788946758959895,-13.694101015877054), linewidth(0.4)); /* dots and labels */ dot((-9.148785977635322,4.836789707219564),dotstyle); label("$A$", (-9.019359834083662,5.178332989794058), NE * labelscalefactor); dot((-14.36894368356669,-3.6587610691001062),dotstyle); label("$B$", (-14.237497038193618,-3.3227382169015836), NE * labelscalefactor); dot((4.635159860248812,-8.435375963416387),dotstyle); label("$C$", (4.768962976776417,-8.091631820657675), NE * labelscalefactor); dot((-1.8992281791900671,5.759993619350055),linewidth(4pt) + dotstyle); label("$D$", (-1.7623478283678313,6.0422629904745095), NE * labelscalefactor); dot((-8.377666341117513,-1.21850999526025),linewidth(4pt) + dotstyle); label("$I$", (-8.224544233457642,-0.9382914150235376), NE * labelscalefactor); dot((-11.347957364959688,-4.418075404118023),linewidth(4pt) + dotstyle); label("$E$", (-11.196463435798412,-4.152111017554817), NE * labelscalefactor); dot((-13.862854380121021,-7.127095305521932),linewidth(4pt) + dotstyle); label("$F$", (-13.719139037785345,-6.847572619677825), NE * labelscalefactor); dot((-11.701507044281756,-1.64178860421456),linewidth(4pt) + dotstyle); label("$P$", (-11.576592636097812,-1.3529778153501544), NE * labelscalefactor); dot((-4.866891911658938,-6.047068516258246),linewidth(4pt) + dotstyle); label("$M$", (-4.7342670307086,-5.776299418834066), NE * labelscalefactor); dot((-18.798087102726157,-2.5455113766722337),linewidth(4pt) + dotstyle); label("$T$", (-18.660818641677555,-2.286022216085042), NE * labelscalefactor); dot((-6.788946758959895,-13.694101015877054),linewidth(4pt) + dotstyle); label("$M'$", (-6.634913032205604,-13.413440624849256), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Nice geo

Let $T$ be the intersection of $BC$ with the line perpendicular to $FE$ at $F$ , $M$ be the midpoint of $BC$ and $M'$ the intersection of $DM$ and $AI$ .
Since $\angle BFE = \angle BFD = \angle DFC =\angle EFC$ , we know that $(T,E;B,C)=-1$ .
Because of this we know that $FE.ED=BE.EC=TE.EM$ , thus $FTDM$ is cylic.
Then we have that $\angle ETF = \angle MTF = \angle FDM = \angle FDM' = \angle FAM' = \angle FAI = \angle FPE$ , from which we conclude that $TPEF$ is cyclic.
From here we know that $\angle TPE = 90$ .
Bearing in mind that $(T,E;B,C)=-1$ and combining it with $\angle TPE = 90$ we conclude that $\angle BPE = \angle EPC$ .

Bonus

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#3 h Apr 21, 2022, 5:37 PM • 4 Y

Y by parmenides51, Mango247, Mango247, Mango247

Iran TST 2012/2

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