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Source: USEMO 2019/1

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dchenmathcounts

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dchenmathcounts

#1 h May 23, 2020, 11:00 PM • 5 Y

Y by Purple_Planet, samrocksnature, HWenslawski, Rounak_iitr, Tastymooncake2

Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$ ). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son

This post has been edited 2 times. Last edited by v_Enhance, Oct 25, 2020, 6:01 AM
Reason: backdate

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1051 posts

#2 h May 23, 2020, 11:00 PM • 7 Y

Y by smartninja2000, Purple_Planet, HUNTER963, AllanTian, samrocksnature, Rounak_iitr, Tastymooncake2

We define point $P$ to be the antipode of $D$ with respect to $(DBEF),$ $G$ as the intersection of lines $EF,DO,$ and $AC,$ and $K$ as the intersection of $AC$ and $BD.$

Claim: $DCGF$ is cyclic.

Proof: $\measuredangle DFG = \measuredangle DFE = \measuredangle DBE = \measuredangle DBA = \measuredangle DCA = \measuredangle DCG.$

Claim: $ADGE$ is cyclic.

Proof: $\measuredangle BAD = \measuredangle BCD = \measuredangle FCD = \measuredangle FGD = \measuredangle EGD.$

Claim: $\measuredangle DKC = 90.$

Proof: Notice that since $DP$ is a diameter of $(DBEPF),$ $\measuredangle DBP = 90.$ But also note that $\measuredangle BPD = \measuredangle BED - \measuredangle AGD \implies AG \parallel BP.$ Now $\measuredangle DKG = \measuredangle DBP = 90,$ so $\measuredangle DKC = 90,$ as desired.

Now, we wish to prove that both $\measuredangle HFE = \measuredangle ACB$ and $\measuredangle HEF = \measuredangle CAB,$ which will prove the desired result by AA similarity. Note that $\measuredangle HFE = \measuredangle DFE - \measuredangle DFH = \measuredangle DFE - (90 - \measuredangle EDF) = 90 - \measuredangle FED = 90 - \measuredangle FBD = 90 - \measuredangle CBK = \measuredangle ACB.$ Also $\measuredangle HEF=\measuredangle DEF -\measuredangle DEH = \measuredangle DEF- (90 - \measuredangle FDE) = 90 - \measuredangle EFD = \measuredangle PFD - \measuredangle EFD = \measuredangle PFE = \measuredangle PDE = \measuredangle GDE = \measuredangle GAE = \measuredangle CAB,$ so $\triangle ABC \sim \triangle EHF,$ as desired.

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2726 posts

#3 h May 23, 2020, 11:01 PM • 6 Y

Y by cosmicgenius, samrocksnature, megarnie, Mango247, Mango247, Tastymooncake2

d michaelpoint

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#4 h May 23, 2020, 11:04 PM • 5 Y

Y by TheUltimate123, zuss77, samrocksnature, PRMOisTheHardestExam, Tastymooncake2

Dumb.

Let concurrence point be $T$ . Since $\triangle DAC\sim\triangle DEF$ , cyclic quads $AETD$ , $CFTD$ by spiral sim lemma.
If $H_D$ is reflection of $H$ over $\overline{EF}$ , then $\measuredangle BDC=\measuredangle EDO=\measuredangle H_DDF$ , so $\triangle DAC\cup B\sim\triangle DEF\cup H_D$ end proof.

This post has been edited 4 times. Last edited by Asuboptimal, May 24, 2020, 8:48 AM

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#5 h May 23, 2020, 11:11 PM • 5 Y

Y by Purple_Planet, samrocksnature, aopsuser305, PRMOisTheHardestExam, Tastymooncake2

Oops, is this the official thread?

Here is an amusing solution I found when testsolving.

Solution

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dchenmathcounts

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dchenmathcounts

#6 h May 23, 2020, 11:15 PM • 2 Y

Y by Purple_Planet, samrocksnature

We claim that $AC\perp BD.$

Let $D'$ be the antipode of $D$ with respect to the circle centered at $O,$ let $P$ be the concurrence point, and let $Q$ be the intersection of $AC$ with $BD.$ It then suffices to show that $\angle BD'D=\angle QPD.$

Note that $\angle QPD=\angle CPD.$ We further claim that $\triangle BED\sim \triangle CPD.$

By the Spiral Similarity Lemma, the center of the spiral similarity that sends $B\to C$ and $E\to P$ is the intersection of $(BEF)$ and $(CPF).$ We claim this point is $D.$ Obviously $D$ lies on $(BEF),$ and also, since $\angle PCD=\angle ACD=\angle ABD=\angle EBD=\angle EFD=\angle PFD,$ $D$ lies on $(CPF)$ as well. So $\triangle BED\sim\triangle CPD.$

Since $\angle BD'D=\angle MOD$ and $\angle BD'D=\angle BED=\angle CPD=\angle QPD,$ $OM\parallel PQ,$ implying $AC\perp BD.$

Now it's well-known that
$\angle FEH=90^{\circ}-\angle EFD$ $\angle EHF=180^{\circ}-\angle FDE$ $\angle HFE=90^{\circ}-\angle DEF.$
Now notice
$\angle BAC=90^{\circ}-\angle ABD=90^{\circ}-\angle EBD=90^{\circ}-\angle EFD=\angle FEH$ $\angle ABC=\angle EBF=180^{\circ}-\angle EDF=\angle EHF$ $\angle BCA=90^{\circ}-\angle DBC=90^{\circ}-\angle DBF=90^{\circ}-\angle DEF=\angle EFH,$ so $\triangle ABC\sim\triangle EHF.$

This post has been edited 1 time. Last edited by dchenmathcounts, May 23, 2020, 11:16 PM

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1487 posts

#7 h May 23, 2020, 11:31 PM • 3 Y

Y by Kagebaka, Purple_Planet, samrocksnature

oh cool, new thread
WLOG, let $C$ be inside $(DEF)$ and $A$ be on the outside.
Let the concurrency point be $T$ .
Note that $D$ is the Miquel point of $BCTE$ , thus $T$ lies on $(AED)$ and $(CFD)$ . The problem falls to simple angle chasing from here, using the fact that $DO$ and $DH$ are isogonal (note: all angles are directed)
$\angle CAB=\angle TAE=\angle TDE=\angle FDH=90^{\circ}-\angle EFD=\angle HEF$ $\angle BCA=\angle FCT=\angle FDT=\angle HDE=90^{\circ}-\angle DEF=\angle EFH$ and we're done.

This post has been edited 1 time. Last edited by franchester, May 23, 2020, 11:31 PM

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#8 h May 23, 2020, 11:37 PM • 3 Y

Y by Purple_Planet, samrocksnature, Mango247

sketch:

Angle chase to reduce to proving $AC\perp BD$ . Note $D$ is the center of the spiral similarity sending $AC$ to $EF$ , which also sends $N$ to $O$ . So Complex bash finishes (set $d=0, n=1, e=az, f=cz, o=z$ , and then prove $NO||AC$ ).

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#9 h May 23, 2020, 11:42 PM • 6 Y

Y by Purple_Planet, smartninja2000, crazyeyemoody907, samrocksnature, Mango247, Tastymooncake2

"Easier" way for people who lack a geometer's brain: 1/2 line of angle chasing + 4 pages of complex bash.

This post has been edited 1 time. Last edited by nukelauncher, May 23, 2020, 11:43 PM

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MP8148

#10 h May 23, 2020, 11:50 PM • 10 Y

Y by Purple_Planet, Smkh, Aryan-23, amar_04, samrocksnature, PRMOisTheHardestExam, Mango247, Mango247, Mango247, Tastymooncake2

No one bothers to make a diagram

$[asy] size(8cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.35)); dotfactor *=1.5; pair O = origin, D = dir(25), F = dir(120), B = dir(200), E = dir(340), G = extension(D,O,E,F), A = intersectionpoint(circumcircle(D,E,G),B--E+dir(E--B)*0.0069), C = extension(A,G,B,F), H = orthocenter(D,E,F); draw(circumcircle(A,D,E)^^unitcircle, purple); draw(A--B--C--A^^E--F--H--E, heavygreen); draw(A--E--D--F); draw(D--O, orange); dot("$A$", A, dir(255)); dot("$B$", B, dir(200)); dot("$C$", C, dir(90)); dot("$D$", D, dir(45)); dot("$E$", E, dir(320)); dot("$F$", F, dir(140)); dot("$G$", G, dir(245)); dot("$H$", H, dir(30)); dot("$O$", O, dir(200)); [/asy]$
Let $G = \overline{BF} \cap \overline{AC}$ . It follows that $D$ is the Miquel point of $ABFG$ , so $AEDG$ is cyclic. Then $\measuredangle EHF = \measuredangle FDE = \measuredangle FBE = \measuredangle CBA$ and $\measuredangle HEF = 90^\circ + \measuredangle DFE = \measuredangle ODE = \measuredangle GDE = \measuredangle GAB = \measuredangle CAB,$ so $\triangle ABC \overset{-}\sim \triangle EHF$ as desired. $\blacksquare$

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924 posts

#12 h May 24, 2020, 3:34 AM • 7 Y

Y by Williamgolly, Purple_Planet, amar_04, srijonrick, samrocksnature, Rounak_iitr, vangelis

Nothing but just trivial angle chase.

Let $X$ be the concurrency point. Notice that since $\angle XFD=\angle EFD=\angle EBD=\angle ACD$ hence $(XFDC)$ is cyclic. Now just notice that $\angle ACB=\angle XCF=\angle XDF=\angle ODF=90^\circ-\angle DBF=90^\circ-\angle DBC$ hence $\overline{AC} \perp \overline{BD}$ . To finish notice that we have that $\angle EFH=90^\circ-\angle DEF=90^\circ-\angle DBF=\angle ACB$ and also we have that $\angle EHF=180^\circ-\angle EDF=\angle EBF=\angle ABC$ and hence the similiarity follows $\qquad \blacksquare$

$[asy] size(8cm); pair A=(-8.5,3.69); pair B=(-4.1004880575605736,6.553349874699234); pair C=(4.46,-2.71); pair D=(-9.186044509110761,-3.766637297304822); pair O=(-5.862865536958581,1.0087848755370215); pair E=(-11.646182892013208,1.642358691716196); pair F=(-0.19673893348792193,2.3290820857385617); pair H=(-9.303235260694732,-1.8127662709240977); pair X=(-5.14334178555926,2.032391005214449); draw(A--B--C--D--A,purple); draw(circumcircle(A,B,C),red); draw(A--E,purple); draw(A--C,green); draw(B--D,green); draw(E--F,blue+dotted); draw(D--X,pink); draw(E--D,blue+white); draw(D--F,blue+white); draw(circumcircle(E,D,F),orange+dotted); draw(circumcircle(D,X,F),cyan+dashed); dot("$A$",A,N); dot("$B$",B,N); dot("$C$",C,SE); dot("$D$",D,W); dot("$E$",E,W); dot("$F$",F,NE); dot("$X$",X,N); dot("$O$",O,S); dot("$H$",H,N); [/asy]$

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#13 h May 24, 2020, 8:46 AM • 3 Y

Y by samrocksnature, Mango247, Aspiring_Mathletes

ez angle chasing

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#14 h May 24, 2020, 8:57 AM • 3 Y

Y by Aspiring_Mathletes, samrocksnature, Mango247

Let $X=EF\cap AC\cap DO$

Claim: $AC\perp BD$
$proof:\;\angle DEF=\angle DBF=\angle DBC=\angle DAC$ which implies $AEXD$ is cyclic.
Hence, we see $\angle BDC=\angle BAC=\angle EAX=\angle EDX$ and so, $\angle XDC=\angle EDB=\angle EFB$ and hence, $DXCF$ is cyclic.

$\therefore 90^{\circ}-\angle CBD=90^{\circ}-\angle FED=\angle XDF=\angle ACB$ and so, $AC\perp BD$

It is now easy to see that $\angle FEH=90^{\circ}-\angle XFD=90^{\circ}-\angle EBD=\angle BAC$ and $EFH=90^{\circ}-\angle FED=90^{\circ}-\angle XDF=90^{\circ}-\angle BCA$ and hence $\triangle ABC \sim \triangle EHF$

$[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -10.951719144867154, xmax = 10.205305648521268, ymin = -8.086246158049283, ymax = 4.260861279967215; /* image dimensions */ /* draw figures */ draw(circle((-0.8189320936592647,-0.10041731422676856), 3.59681270586667), linewidth(0.5)); draw((1.1309254832320157,2.9220183047606043)--(2.31409603169182,-1.8671081210299814), linewidth(0.5)); draw((-2.781184345331607,2.913986066525506)--(1.1309254832320157,2.9220183047606043), linewidth(0.5)); draw((-2.781184345331607,2.913986066525506)--(2.31409603169182,-1.8671081210299814), linewidth(0.5)); draw((-2.781184345331607,2.913986066525506)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); draw((-3.7113245978215597,-2.238436781619356)--(2.31409603169182,-1.8671081210299814), linewidth(0.5)); draw(circle((0.17674570293097336,-1.034699428989725), 4.070144257604352), linewidth(0.5)); draw((2.31409603169182,-1.8671081210299814)--(2.8637276646324716,-4.091855331981989), linewidth(0.5)); draw((2.8637276646324716,-4.091855331981989)--(-0.793673616147727,2.9180667698726), linewidth(0.5)); draw((-0.793673616147727,2.9180667698726)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); draw((-3.7113245978215597,-2.238436781619356)--(2.8637276646324716,-4.091855331981989), linewidth(0.5)); draw((-0.793673616147727,2.9180667698726)--(-1.9947619551987608,-1.3428264857492949), linewidth(0.5)); draw((-1.9947619551987608,-1.3428264857492949)--(2.8637276646324716,-4.091855331981989), linewidth(0.5)); draw((-1.9947619551987608,-1.3428264857492949)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); draw(circle((9.950552748009656,16.239095787241443), 3.056934751946367), linewidth(0.5)); draw(circle((13.43307492051343,13.410111137651443), 5.583923634788888), linewidth(0.5)); draw((xmin, -3.9076923076922974*xmin + 64.55197035721586)--(xmax, -3.9076923076922974*xmax + 64.55197035721586), linewidth(0.5)); /* line */ draw((xmin, 0.10780669144981424*xmin + 17.462742100974747)--(xmax, 0.10780669144981424*xmax + 17.462742100974747), linewidth(0.5)); /* line */ draw((7.68479475445666,18.291214397923607)--(7.879417498909836,13.990707729633588), linewidth(0.5)); draw((7.879417498909836,13.990707729633588)--(12.703577624479204,14.910297793866512), linewidth(0.5)); draw((13.966860208217316,18.968463089964718)--(7.879417498909836,13.990707729633588), linewidth(0.5)); draw((7.879417498909836,13.990707729633588)--(14.490580802889738,7.927239219769952), linewidth(0.5)); draw((13.966860208217316,18.968463089964718)--(14.490580802889738,7.927239219769952), linewidth(0.5)); draw((11.726868383307151,18.726976982446523)--(7.879417498909836,13.990707729633588), linewidth(0.5)); draw((1.1309254832320157,2.9220183047606043)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); draw((1.116816708577481,-0.7436556730674665)--(-3.7113245978215597,-2.238436781619356), linewidth(0.5)); draw(circle((-0.5930579300844421,-3.7655985655679327), 3.472147768381024), linewidth(0.5)); draw(circle((-1.7815925087753384,0.07336679239028558), 3.0113621004951305), linewidth(0.5)); /* dots and labels */ dot((-2.781184345331607,2.913986066525506),dotstyle); label("$A$", (-2.7203141861894706,3.087307560958954), NE * labelscalefactor); dot((1.1309254832320157,2.9220183047606043),dotstyle); label("$B$", (1.1970411857113545,3.087307560958954), NE * labelscalefactor); dot((2.31409603169182,-1.8671081210299814),dotstyle); label("$C$", (2.387123830339453,-1.7060808687931), E * labelscalefactor); dot((-3.7113245978215597,-2.238436781619356),linewidth(4pt) + dotstyle); label("$D$", (-3.8773389795778996,-2.598642852264172), W * labelscalefactor); dot((0.17674570293097336,-1.034699428989725),dotstyle); label("$O$", (0.12266102042209867,-0.8135188853220279), NE * labelscalefactor); dot((-0.793673616147727,2.9180667698726),linewidth(4pt) + dotstyle); label("$E$", (-0.7203141861894713,3.0542497097192847), NE * labelscalefactor); dot((2.8637276646324716,-4.091855331981989),linewidth(4pt) + dotstyle); label("$F$", (2.9325783757939985,-3.954014753090615), SE * labelscalefactor); dot((-1.9947619551987608,-1.3428264857492949),linewidth(4pt) + dotstyle); label("$H$", (-1.9269257564374047,-1.2102131001980598), NE * labelscalefactor); dot((7.68479475445666,18.291214397923607),dotstyle); label("$G$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((11.726868383307151,18.726976982446523),dotstyle); label("$I$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((12.703577624479204,14.910297793866512),dotstyle); label("$J$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((7.879417498909836,13.990707729633588),dotstyle); label("$K$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((13.43307492051343,13.410111137651443),dotstyle); label("$L$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((14.490580802889738,7.927239219769952),linewidth(4pt) + dotstyle); label("$M$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((13.966860208217316,18.968463089964718),linewidth(4pt) + dotstyle); label("$N$", (-10.951719144867154,4.260861279967216), NE * labelscalefactor); dot((1.116816708577481,-0.7436556730674665),linewidth(4pt) + dotstyle); label("$X$", (1.1805122600915197,-0.6151717778840118), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

This post has been edited 3 times. Last edited by Greenleaf5002, May 24, 2020, 9:11 AM
Reason: i want to kill myself

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130 posts

#15 h May 24, 2020, 9:00 AM • 2 Y

Y by samrocksnature, Mango247

redacted

This post has been edited 3 times. Last edited by Greenleaf5002, May 24, 2020, 9:10 AM

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#16 h May 24, 2020, 9:03 AM • 4 Y

Y by mkomisarova, samrocksnature, Mango247, Mango247

Let $X$ be intersection of $AC$ , $DO$ , $EF$ .

Claim: Quadrilaterals $XCFD$ and $AEXD$ are cyclic.
Proof: Note that from cyclic quadrilaterals $EBFD$ and $ABCD$ we obtain:
$\angle EFD = \angle EBD = \angle ABD = \angle ACD = \angle XCD$ $\angle DEF = \angle DBF = \angle DBC = \angle DAC = \angle DAX$ This proofs our claim.

Since $O$ is circumcenter of $\triangle DEF$ , we get that $\angle ODF = 90 - \angle FED$ . Since $XCFD$ is cyclic:
$\angle ODF = \angle XDF = \angle BCX = \angle BCA = 90 - \angle FED$ But since $H$ is orthocenter of $\triangle DEF$ , we get that $\angle HFE = 90 - \angle FED = \angle BCA$ . Similarly we can prove that $\angle BAC = \angle HEF$ , which implies that $\triangle ABC \sim \triangle EHF$ $ and we are done.

This post has been edited 1 time. Last edited by Kimchiks926, May 24, 2020, 9:11 AM
Reason: typo

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Mathematicsislovely

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Mathematicsislovely

#18 h May 24, 2020, 10:48 AM • 1 Y

Y by samrocksnature

As first, as $H$ is the orthocentre of $DEF$ implies $D$ is the orthocentre of $HEF$ .So $\angle EHF=180^{\circ}-\angle EDF$

Claim:
$\angle ABC=180^{\circ}-\angle EDF(=\angle EHF)$
proof:
$\angle DFE=\angle DBE=\angle DBA=\angle DCA$ [by 2 cyclic quadrilateral].Similarly , $\angle DEF=\angle DAC$ .So
$\angle EDF=\angle ADC=180^{\circ}-\angle ABC$ .
SO, $\angle ABC=180^{\circ}-\angle EDF(=\angle EHF)$ $\blacksquare$

Let $AC,WE,DO$ concur at $X$ .From the proof of last claim we have that
$\angle DFX=\angle DFE=\angle DBE=\angle DBA=\angle DCA=\angle DCX$ .So, $XDFC$ concyclic.So $\angle XDF= \angle XCB=\angle ACB=x$ .

NOW,as $O$ is the centre of $(BDF)$ So $\angle DOF=180^{\circ}-2\angle ODF=180^{\circ}-2x$ which implies $\angle DBF=90^{\circ}-x$ .

Again as $DXEA$ cyclic,
So $90^{\circ}-x=\angle DBC=\angle DAX=\angle DEX$ .As $D$ is orthocentre of $HEF$ so
$\angle HFE=90^{\circ}-\angle DEF=x=\angle ACB$ .So $ABC$ and $EHF$ are similar. $\blacksquare$

This post has been edited 5 times. Last edited by Mathematicsislovely, May 24, 2020, 10:56 AM

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6858 posts

#19 h May 24, 2020, 12:47 PM • 3 Y

Y by v4913, samrocksnature, Mango247

Also some solutions posted in https://artofproblemsolving.com/community/c5h2119379_concurrence_implies_similarity_on_cyclic_quads.

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#20 h May 24, 2020, 1:47 PM • 1 Y

Y by samrocksnature

>

forgot about directed angles so i think i only got a 6

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1201 posts

#21 h May 24, 2020, 1:49 PM • 4 Y

Y by Mathscienceclass, cocohearts, samrocksnature, Mango247

i was thinking during the test: "make sure to use directed angles"
me during the writeup: "normal angles go brrbrr"

This post has been edited 1 time. Last edited by brianzjk, May 24, 2020, 1:50 PM

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#22 h May 24, 2020, 3:21 PM • 3 Y

Y by v4913, samrocksnature, Mango247

I didn’t solve this live ~~doing math the day after you get a concussion is hard~~. Also not sure if this is right because I didn’t direct anything. Define $X$ to be the intersection of $EF, AC, DO$ .
Claim: $AXDE$ is cyclic.
Proof. Since $ABCD$ is cyclic, $\angle DAC = \angle DBC = \angle DBF$ . Since $EBFD$ is cyclic, $\angle DBF = \angle DEF$ . Thus, $\angle DAC = \angle DEF$ which proves that $AXDE$ is cyclic.
Claim: $XFCD$ is cyclic.
Proof. $\angle EAD = \angle BCD$ since $ABCD$ is cyclic. Furthermore, $\angle EAD = \angle EXD = 180 - \angle DXF$ . $\angle DXF = 180 - \angle EAD$ . Hence, $\angle BCD + \angle DXF = 180$ , proving that $XFCD$ is cyclic.
Claim: $BD \perp AC$ .
Note that $O$ is the center of the circumcircle of $BDF$ . Therefore, $\angle DOF = 2 \angle BDF$ , which implies that $\angle XDF = 90 - \angle DBF$ . Since $XFCD$ is cyclic, $\angle XDF = \angle XCF = 90 - \angle DBF$ . Hence, $BD \perp AC$ .
Claim: $\triangle ABC \sim \triangle EHF$ .
Let $\angle FED = a$ and $\angle ABD = b$ . Then, since $EAXD$ is cyclic, $\angle DAX = a$ and since $ABCD$ is cyclic, $\angle DBC = a$ as well. Since $BD \perp AC$ , $\angle XCF = 90 - a$ . Similarly, since $ABCD$ is cyclic, $\angle ACD = b$ and since $BD \perp AC$ , $\angle BAC = 90 - b$ . As $XFCD$ is cyclic, $\angle XFD = \angle XCD = b$ . Thus, $\angle ABC = b + a, \angle BCA = 90 -a, \angle BAC = 90 - b$ and $\angle DEF = a, \angle EFD = b, \angle EDF = 180 - a - b$ . Since $EH \perp DF$ , $\angle EHF = 90 - b$ and since $HF \perp ED$ , $\angle EFH = 90 -a$ , which implies that $\angle EHF = a + b$ . This proves that $ABC$ and $EHF$ are similar.

EDIT: 1500th post

This post has been edited 1 time. Last edited by AopsUser101, May 24, 2020, 3:23 PM

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1218 posts

#23 h Jun 11, 2020, 7:25 PM • 3 Y

Y by amar_04, itslumi, samrocksnature

Solution

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1010 posts

mira74

#24 h Jun 18, 2020, 1:29 AM • 2 Y

Y by tworigami, Mango247

Didn't post my solution at the time but i thought it was interesting and I don't see it so

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.7360599458428776, xmax = 1.7777171367820195, ymin = -1.8136357022133056, ymax = 1.1697599339776357; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-0.37591727958583643,0.9266532247333865)--(-0.8831267030065891,-0.4691345504614979)--(0.8946747462966459,-0.44671814194079057)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw(circle((0,0), 1), linewidth(2) + wrwrwr); draw((-0.37591727958583643,0.9266532247333865)--(-0.8831267030065891,-0.4691345504614979), linewidth(2) + rvwvcq); draw((-0.8831267030065891,-0.4691345504614979)--(0.8946747462966459,-0.44671814194079057), linewidth(2) + rvwvcq); draw((0.8946747462966459,-0.44671814194079057)--(-0.37591727958583643,0.9266532247333865), linewidth(2) + rvwvcq); draw((0,0)--(0.6657366321263676,-0.7461867974207541), linewidth(2) + wrwrwr); draw((-0.7913909810231188,-0.21668734940272177)--(0.938374295807192,-0.4939525876257939), linewidth(2) + wrwrwr); draw((-0.37591727958583643,0.9266532247333865)--(-1.2519448752529072,-1.484083942803701), linewidth(2) + wrwrwr); draw((-1.2519448752529072,-1.484083942803701)--(0.7189832312577051,-0.2568147809823426), linewidth(2) + wrwrwr); draw((-0.37591727958583643,0.9266532247333865)--(0.938374295807192,-0.4939525876257939), linewidth(2) + wrwrwr); /* dots and labels */ dot((-0.37591727958583643,0.9266532247333865),dotstyle); label("$B$", (-0.4067914457177986,0.9808115436855426), NE * labelscalefactor); dot((-0.8831267030065891,-0.4691345504614979),dotstyle); label("$E$", (-0.9603770804332306,-0.46779278188717005), NE * labelscalefactor); dot((0.8946747462966459,-0.44671814194079057),dotstyle); label("$F$", (0.8628091416834613,-0.37497602876122965), NE * labelscalefactor); dot((0.6657366321263676,-0.7461867974207541),dotstyle); label("$D$", (0.6838054035120044,-0.8490044465115681), NE * labelscalefactor); dot((0,0),dotstyle); label("$O$", (0.014198827389146598,0.03275470818486565), NE * labelscalefactor); dot((0.4040742706873469,-0.45290415370606474),linewidth(4pt) + dotstyle); label("$T$", (0.3556318835309998,-0.5374053467316253), NE * labelscalefactor); dot((0.6619976062882014,-0.44965198240194176),linewidth(4pt) + dotstyle); label("$P$", (0.6738607513913678,-0.42469928936441204), NE * labelscalefactor); dot((-0.7913909810231188,-0.21668734940272177),linewidth(4pt) + dotstyle); label("$Q$", (-0.7780584582215614,-0.18934252250934888), NE * labelscalefactor); dot((0.938374295807192,-0.4939525876257939),linewidth(4pt) + dotstyle); label("$R$", (0.9523110107691899,-0.46779278188717005), NE * labelscalefactor); dot((0.4610298387881029,-0.41743872394575415),linewidth(4pt) + dotstyle); label("$S$", (0.4749677089786378,-0.39155044896229046), NE * labelscalefactor); dot((-1.2519448752529072,-1.484083942803701),linewidth(4pt) + dotstyle); label("$A$", (-1.2388273398110525,-1.458943109910605), NE * labelscalefactor); dot((0.7189832312577051,-0.2568147809823426),linewidth(4pt) + dotstyle); label("$C$", (0.7335286641151868,-0.22912113099189477), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Screw directed angles.

Draw the simson line of $D$ to both $BAC$ and $BEF$ as labelled. Now, $\angle ODS = \angle TDS = \angle TPS = \angle EPQ = \angle EDQ = 90^{\circ} - \angle DEQ = 90^{\circ} - \angle DEB = \angle ODB$ so $D$ , $S$ , $B$ are collinear. Thus, $DB \perp AC$ , implying the result.

Edit: Here's another solution I'm surprised not to see:

Note that since $\angle FDC = \angle EDA$ by spiral similarity, $DC$ and $DE$ are isogonal in $\angle ADF$ . Thus, by the second isogonality lemma, $DO$ , $DB$ are isogonal in $\angle ADF$ , giving $\angle ADB = \angle FDO = 90^{\circ} - \angle DBF = 90^{\circ} - \angle DBC = 90^\circ - \angle DAC$ giving $BD \perp AC$ , implying the result.

This post has been edited 2 times. Last edited by mira74, Jun 18, 2020, 1:43 AM

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757 posts

#25 h Oct 21, 2020, 6:20 AM

Y by

This was a very nice problem.

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crazyeyemoody907

450 posts

crazyeyemoody907

#26 h Jan 29, 2021, 10:07 PM

Y by

nukelauncher wrote:

"Easier" way for people who lack a geometer's brain: 1/2 line of angle chasing + 4 pages of complex bash.

Me neither... can't launch nwooks during live olympiad

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54 posts

#27 h Mar 6, 2021, 10:20 AM

Y by

Let $J$ be the intersection of $EF$ and $AC$ , note that $D$ is the Miquel point of $\mathcal{Q}(AB, BC, AC, EF)$ , hence $\measuredangle EDJ = \measuredangle BDC = \measuredangle BAC$ . Since $D, O, J$ are collinear, $\measuredangle EDJ = \measuredangle HDF = \measuredangle FEH$ . Combining that $\measuredangle ABC = \measuredangle EBD = \measuredangle EDF = \measuredangle FHE$ , we have $\triangle ABC \sim \triangle EHF$ .

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1522 posts

#28 h Apr 28, 2021, 8:15 PM • 2 Y

Y by centslordm, Mango247

Click to reveal hidden text

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#29 h Jul 14, 2021, 2:08 AM • 1 Y

Y by centslordm

Let $X$ be the concurrency point. Observe that
$\measuredangle DEX=\measuredangle DEF=\measuredangle DBF=\measuredangle DBC=\measuredangle DAC=\measuredangle DAX,$ hence $ADEX$ is cyclic. Likewise, $CDFX$ is cyclic.
Now let $M$ be the midpoint of $\overline{ED}$ , and $P$ be the intersection between $\overline{AC}$ and $\overline{BD}$ . Since $\overline{XD}$ and $\overline{XE}$ are radii of the same circle, we have $\overline{XM} \perp \overline{DE}$ . Since we have
$\measuredangle BAP=\measuredangle EAX=\measuredangle EDX=\measuredangle MDX,$ as well as
$\measuredangle PBA+\measuredangle DBA=\measuredangle DCA=\measuredangle DCX=\measuredangle DFX=\measuredangle DFE=\measuredangle DXM,$ it follows that $\triangle ABP\sim \triangle DXM$ . As $\angle XMD=90^\circ$ , we have $\angle APB=90^\circ$ , hence $\overline{AC} \perp \overline{BD}$ .
Since $\overline{DF} \perp \overline{EH}$ , we have $\measuredangle FEH=90^\circ-\measuredangle DFE$ . Similarly, $\angle EFH=90^\circ-\angle DEF$ . Now, we have
$\measuredangle DFE=\measuredangle DBE=\measuredangle PBA=90^\circ-\measuredangle BAP,$ so $\measuredangle FEH=\measuredangle BAP=\measuredangle BAC$ . Likewise, $\measuredangle EFH=\measuredangle BCA$ , so $\triangle EHF$ and $\triangle ABC$ are similar, as desired. $\blacksquare$

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#30 h Aug 16, 2021, 5:04 PM

Y by

Cool problem but somebody should get me glasses.

Let $X$ be the concurrence point.
Notice that $\triangle DFE \sim \triangle DCA$ Now, $\angle HFE = 90^{\circ}-\angle DEF = 90^{\circ} - \angle DAC$ Consequently, we have to prove that $\angle ACB = 90^{\circ} - \angle CAD$ because then $\angle HFE = \angle ACB$ and analogously $\angle HEF = \angle BAC$ Notice that $\angle DFX = \angle DFE = \angle DCA$ meaning that $DCXF$ is cyclic.
Then $\angle ACB = \angle FCX = \angle FDX = \angle FDO = 90^{\circ}-\angle DEF = 90^{\circ} - \angle DBC$ as desired. $\blacksquare$

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484 posts

554183

#31 h Sep 22, 2021, 1:53 PM

Y by

For the given condition to hold, it is enough to show that $AC \perp BD$ .
Let the concurrence point be $X$ . Note that
$\angle{XFD}=\angle{EFD}=\angle{EBD}=\angle{ABD}$ This shows that $XCFD$ is cyclic.
Now, notice
$\angle{XDC}=\angle{XFC}=\angle{XFB}=\angle{EFB}=\angle{EDB}$ Notice
$\angle{DOE}=2\angle{DFE}$ Therefore,
$\angle{XDE}=90-\angle{DFE} \implies \angle{BDX}=90-\angle{DFE}-\angle{EDB} \implies \angle{BDC}=90-\angle{DFX}=90-\angle{DBA}$ As desired

This post has been edited 1 time. Last edited by 554183, Sep 22, 2021, 1:54 PM
Reason: Lol

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primesarespecial

364 posts

primesarespecial

#32 h Sep 22, 2021, 2:47 PM

Y by

Let $Q$ be the concurrence point.
We see that $\angle FHE=\angle ABC$ .
Thus it suffices to prove $AC \perp BD$ .
Now, we see that $\angle QFD=\angle EBD=\angle ACD$ .
Thus $QFCD$ is cyclic.
Which gives that $\angle BCA=\angle FCQ=\angle FDO=90- \angle FED=90-\angle DBC$
which gives the desired.

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594 posts

#33 h Sep 25, 2021, 12:00 PM

Y by

(Sorry if I mess up dangles) Let $X$ be the point of concurrency.
$\measuredangle DAX = \measuredangle DBC = \measuredangle DBF = \measuredangle DEX \implies DAEX \text{ is cyclic.}$ $\measuredangle DCX = \measuredangle DCA = \measuredangle DBE = \measuredangle DFX \implies DFCX \text{ is cyclic.}$ Note that $\{EH , EO\}$ and $\{FH,FO\}$ are isogonal wrt $\triangle EDF$ .
$\measuredangle BAC = \measuredangle EDG = \measuredangle OED = \measuredangle FEH$ $\measuredangle BCA = \measuredangle FDG = \measuredangle OFD = \measuredangle EFH$ which finishes the problem.

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3217 posts

#34 h Jan 10, 2022, 5:18 AM • 1 Y

Y by centslordm

We can see that $D$ is the miquel point of the complete quad $ABCEFX$ , giving us cyclic quad $AEDX$ where $X=EF\cap AC$ .

Note that $\angle XDB=\frac{180-\angle BOD}{2}=90-\angle BED=\angle AXD-90=90-\angle CXD$ (assuming that $E$ lies beyond $A$ ), hence $BD$ is perpendicular to $AC$ . Now, the desired spiral similarity sends $B$ to a point $H'$ on $(DEF)$ such that $DH'\perp EF$ , and we are done by orthocenter-reflection.

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1080 posts

#35 h Jan 21, 2022, 3:05 AM • 5 Y

Y by centslordm, teomihai, megarnie, crazyeyemoody907, Rounak_iitr

Let $X=\overline{AC}\cap\overline{DO}\cap\overline{EF},Y=\overline{AC}\cap\overline{BD},Z=\overline{DO}\cap\overline{BC},$ and $D'$ the reflection of $D$ in $O.$ We see $\angle EHF=\angle 180-\angle EDF=\angle CBA.$ Notice $D$ is the Miquel point of $BEXZ$ so $D$ lies on $(DFX).$ Hence, $\angle YBC=\angle DD'F=90-\angle D'DF=90-\angle YCB$ so $\overline{AC}\perp\overline{BD}.$ Therefore, $\angle ACB=90-\angle CBD=90-\angle FED=\angle HFE.$ $\square$

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85 posts

ALM_04

#36 h Jan 30, 2022, 8:10 AM

Y by

Since, $BEDF$ is a cyclic quadrilateral.
$\measuredangle{FHE}=\measuredangle{EDF}=\measuredangle{EBF}=\measuredangle{ABC}$
Let $X=\overline{EF}\cap\overline{AC}\cap\overline{DO}$ .
$\measuredangle{DCX}=\measuredangle{DCA}=\measuredangle{DBA}=\measuredangle{DBE} =\measuredangle{DFE}=\measuredangle{DFX}\implies DXFC$ is a cyclic quadrilateral.

Now,
Using the above thing and the fact that $O$ is the circumcenter of $DBF$ . It can be shown that $\overline{CA}\perp \overline{BD}$ .
Hence, $\measuredangle{EFH}=90^{\circ}-\measuredangle{DEF}=90^{\circ}-\measuredangle{DBF}=\measuredangle{BCA}$ which proves that $ABC$ and $EHF$ are similar. $\blacksquare$

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140 posts

#37 h Sep 23, 2023, 5:47 PM

Y by

Let the concurrence point be P. We make a couple of observations:
\begin{itemize}
\item DAC=DBF=DEF,ACD=EBD=EFD\implies ACD\sim EFD\implies ABC=180-ADC=180-EDF=EHF
\item DFP=ACD=DCP,DEP=DAC=DAP, so we derive those two cyclic quads.
\item ACB=ODF=90-DBC\implies AC\perp BD
\item EFH=90-DEF=90-DAC=ADB=ACB
\end{itemize}
By AA similarity we get the desired conclusion.

\textbf{Remark.} It's not easy to conjecture AC\perp BD, but it would be wanted from the last item in our list (we'd want 90-DAC=90-DEF=EFH=ACB=ADB).

btw does anyone have recommendations on how to get the itemize thing to work, I've been working on overlefa with \usepackage[s3xy]{evan} which makes it work but idk how to on aops

This post has been edited 1 time. Last edited by signifance, Sep 23, 2023, 5:48 PM

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#38 h Nov 23, 2023, 10:14 PM

Y by

Let $P = \overline{AC} \cap \overline{EF}$ .
Claim: $AEPD$ and $CPDF$ are cyclic.
Proof:
$\angle EPA = \angle CPF = 180^{\circ} - \angle PCF - \angle CFP = \angle BCA - \angle BFE$ $= \angle BDE + \angle EDA - \angle BFE = \angle EDA$ $CPDF$ follows similarly. $\square$

Let $\angle A$ and $\angle B$ denote the angles at $A$ and $B$ of cyclic quadrilateral $ABCD$ . Notice that $\angle EDF = 180^{\circ} - \angle B$ , so $\angle EHF = \angle ABC$ . It suffices to show that $\angle EFH = \angle BCA$ .

Now if we let $\angle BAC = \alpha$ , we get $\angle FED = \angle A - \alpha$ , so $\angle EFH = 90^{\circ} - \angle A + \alpha$ .
Claim:
$\alpha = \frac{90^{\circ}+\angle A - \angle B}{2}$ which finishes since then $\angle BCA = 180^{\circ} - \angle B - \alpha = \angle EFH$ .
Proof:
$\angle EFD = \angle EBD = \angle ABD = \angle ACD = 180^{\circ} - \angle CAD - \angle ADC$ $\implies \angle EFD = 180^{\circ} - (\angle A - \alpha)- (180^{\circ} - \angle B) = \angle B - \angle A + \alpha$ $\implies EDO = 90^{\circ} - \angle EFD = 90+\angle A-\angle B- \alpha$ Since $AEPD$ cyclic, $\alpha = \angle EAP = \angle EDP = \angle EDO$ , which finishes. $\square$

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1247 posts

ezpotd

#39 h Nov 24, 2023, 7:12 AM

Y by

Observe that $D$ is the orthocenter of $EHF$ , so it suffices to prove that $ACX$ and $EFD$ are similar where $X$ is the orthocenter of $ABC$ .

Our main claim is that $D$ is the reflection of $X$ over $AC$ . To see this, animate $O$ linearly along the perpendicular bisector of $BD$ . Then $DEFO$ is similar to a fixed configuration, so $EF \cap DO$ moves linearly.

It is easy to see that when $O$ is the circumcenter of $ABCD$ , we have $EF \cap DO$ lying on $AC$ . If $BD$ is not perpendicular to $AC$ , then we can choose some point $O$ such that the entire line segment $DO$ is parallel to $AC$ , so $EF \cap DO$ does NOT lie on $AC$ , so the line that $EF \cap DO$ moves along is never on $AC$ unless $AC = EF$ , which we can discard. As a result, $BD$ must be perpendicular to $AC$ and $X$ is the desired reflection point.

Now it is easy to see the desired result, observe $ACX$ is similar to $CAD$ , which is always similar to $EFD$ as we animate $O$ along the perpendicular bisector.

This post has been edited 2 times. Last edited by ezpotd, Nov 24, 2023, 6:34 PM

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3430 posts

#40 h Jan 29, 2024, 6:30 AM

Y by

Claim: $\triangle DEF \sim \triangle DAC$ .

Proof: this is true even without the concurrency information. We have
$\angle DEA = \angle DBE = \angle DFB = \angle DFC.$ Furthermore, we have
$\angle EDF = \angle EBF = 180^{\circ} - \angle EBC = \angle ADC,$ so $\angle EDA = \angle FDC$ .

Thus, $\triangle DEA \sim \triangle DFC$ , so $\triangle DEF \sim \triangle DAC$ .

Let $X$ be the concurrency point in the problem, let $D'$ and $D''$ be the antipodes of $D$ WRT $(DEF)$ and $(ABCD)$ , respectively, and let $Y$ be the intersection of $\overline{DG}$ with $\overline{AC}$ .
Because of similarity, $\frac{DY}{DG} = \frac{DX}{DD'}$ , so $\overline{GD'} \parallel \overline{YX}$ .

Because $B$ lies on both $(DAC)$ and $(DEF)$ , we have that $\angle DBG = \angle DBD' = 90^{\circ}$ , hence $D$ , $B$ and $G$ are collinear. It remains to prove that $B \neq G$ ; then, from there, we know that $B$ is the unique point which lies on $(ACD)$ satisfying $\overline{BG} \parallel \overline{AC}$ , which means $B$ is the reflection of the orthocenter of $\triangle ACD$ across $\overline{AC}$ , which means $\triangle ABC \sim \triangle EHF$ as desired.

Proof: Assume, for the sake of contradiction, that $B = G$ ; then, $G$ would lie on $(EDF)$ . Since $\angle DCG = \angle DAG = 90^{\circ}$ , line $AC$ would be the Simson line in $(DEF)$ WRT $D$ and $\triangle BEF$ . So, $X$ would be the foot from $D$ to $\overline{EF}$ , implying $DE = DF$ , implying $DAC$ collinear, which is a contradiction.

remark: dealing with the edge case at the end was by far the most interesting part of this problem

This post has been edited 2 times. Last edited by ihatemath123, Jan 29, 2024, 6:34 AM

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#41 h Mar 10, 2024, 3:48 AM

Y by

The key is the following:

Claim. $\triangle DAE \sim \triangle DCF$ .

Proof. $\angle EAD = \angle FCD$ is evident. Other equality comes from $BEDF$ cyclic. $\blacksquare$

So $\triangle DAC \sim \triangle DEF$ by spiral similarity, ergo for $G = \overline{AC} \cap \overline{EF}$ , $GEAD$ and $GFCD$ are cyclic. So $\angle EFD = \angle ACD$ and $\angle FED = \angle CAD$ . Then it suffices to show that $ABCD$ is orthodiagonal.

To see this, let $D'$ be the $D$ -antipode. It suffices to show that $\angle BDD' + \angle AGD = 90^\circ$ , but this follows as $\angle AGD = \angle AED$ and $\angle BDD' = \angle BED - 90^\circ$ .

This post has been edited 1 time. Last edited by HamstPan38825, Mar 10, 2024, 3:54 AM

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#42 h Jun 2, 2024, 8:42 PM

Y by

Could someone check my solution...

$\text{Let the perpendicular from } D \text{ to } EF \text{ meet the circle centred at }O \text{ again at }H'$
$\text{We have that }\triangle ABC \sim \triangle EHF \cong \triangle EH'F \text{ and so,}$
$\angle H'DF = \angle H'EF = \angle BAC = \angle BDC \text{ with } \angle CBD = \angle FBD = \angle FH'D$ $\implies \text{ a spiral similarity centred at } D \text{ mapping } (ABCD) \text{ to }(EH'FD)$ $\text{We get that }AC \perp BD$
$\text{Now let } AC \cap EF = T; \text{ then } \angle DCA = \angle DFE = \angle DFT \implies (DCTF) \text{ and hence we have }$
$\angle CDT = \angle CFT = \angle BFE = \angle BDC = \angle H'DF = \angle CDO \implies D-O-T \text{ collinear}$ $\text{i.e. AC, DO, EF concurrent }$

This post has been edited 1 time. Last edited by DroneChaudhary, Jun 3, 2024, 1:50 AM
Reason: formatting

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#43 h Jun 3, 2024, 6:16 AM • 1 Y

Y by Rounak_iitr

sobad.

We let the concurrency point be $K$ .

Claim: $DAKE$ cyclic.

By immediate angle chase, we get

$\measuredangle DAC = \measuredangle DBC = \measuredangle DEF = \measuredangle DEK$
implying the claim. Now notice that,

$\measuredangle DKF = \measuredangle DKA + \measuredangle AKF = \measuredangle DEA + \measuredangle EDA = \measuredangle DAB.$
Also,

$\measuredangle KFD = \measuredangle EFD = \measuredangle EBD = \measuredangle ABD.$
So, $\triangle DKF \sim \triangle DAB.$ Now finally as,

$\measuredangle EHF = - \measuredangle EDF = -\measuredangle EBF = \measuredangle EBC = \measuredangle ABC$
and,

$\measuredangle EFH = \measuredangle OFD = \measuredangle ODF = \measuredangle KDF = \measuredangle ADB = \measuredangle ACB$
we get $\triangle EHF \sim \triangle ABC$ as needed.

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792 posts

#44 h Jun 8, 2024, 1:43 AM • 1 Y

Y by Rounak_iitr

Suppose the concurrency point is $K$ and the center of $(ABCD)$ is $P$ . Notice that
$\measuredangle KFD = \measuredangle ABD = \measuredangle ACD,$
so $CDFK$ is cyclic. Then
$\measuredangle POD = \measuredangle CFD = \measuredangle CKD \implies AC \parallel OP \perp BD,$
so our desired similarity is derived from
$\begin{align*} \measuredangle ABC &= \measuredangle EBF = \measuredangle EDF = \measuredangle FHE. \\ \measuredangle CAB &= 90 - \measuredangle ABD = 90 - \measuredangle EFD = \measuredangle HEF. \\ \measuredangle BCA &= 90 - \measuredangle DBC = 90 - \measuredangle DEF = \measuredangle EFH. \quad \blacksquare \end{align*}$

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250 posts

L13832

#45 h Oct 27, 2024, 8:04 AM • 1 Y

Y by Rounak_iitr

Solution

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1299 posts

#46 h Dec 22, 2024, 2:26 AM

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Pure Angle-Chase:

Claim: $DEAX, DCFX$ are cyclic.
Notice that: $\angle ACD = \angle ABD = \angle EFD = \angle XFD$ thus $CXFD$ is cyclic.
Notice that: $\angle AXD=\angle CXD=\angle CFD=\angle BFD=\angle BED = \angle AED$ thus $AEXD$ is cyclic.

Notice that: $\angle CDF = \angle CXE = \angle AXE = \angle ADE$ and thus: $\angle ADC = \angle BDF$ which implies $\angle ABC = \angle EHF$ .

Notice that: $\angle ACB = \angle XCF = \angle XDF = \angle EDH = \angle EFH$ . Similarly, $\angle BAC = \angle HEF$ and thus: $\triangle ABC \sim \triangle EHF$

This post has been edited 2 times. Last edited by Saucepan_man02, Feb 17, 2025, 1:34 PM
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#47 h Mar 16, 2025, 10:25 PM

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nice and chill
let $X$ be the concurrency point
Note that it is given $BDEF$ is cyclic.
Thus, $\angle DEF = \angle DBC = \angle CAD,$ so $AEXD$ is cyclic.
From here, note that $\angle HEF = \angle HDF = \angle EDX = \angle BAX,$ where the second equality is from the isogonality of circumcenter and orthocenter.
Finally, observe that $\angle ABC = 180 - \angle BEF = 180 - \angle EDF = \angle EHF,$ which finishes!

This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 16, 2025, 10:26 PM
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