Prove DK and BC are perpendicular.

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yunxiu

571 posts

yunxiu

#1 h Apr 13, 2012, 12:10 AM • 8 Y

Y by MathAllTheWay, mathematicsy, Quidditch, ys311, Adventure10, ItsBesi, Rounak_iitr, DEKT

Let $ABC$ be a triangle with circumcentre $O$ . The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$ . (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$ . Prove that the lines $DK$ and $BC$ are perpendicular.

Netherlands (Merlijn Staps)

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Mewto55555

4210 posts

Mewto55555

#2 h Apr 13, 2012, 3:06 AM • 5 Y

Y by Aspirant-to-IMO, egxa, Adventure10, The.wrld, and 1 other user

So we'll just consider acute $ABC$ ; cases where $O$ is outside can be done pretty much analogously (or just replace everything with directed angles, which I'm a tad too lazy to do).

We start by angle chasing. $\angle{COB}=2A \implies \angle{OCB}=\angle{OBC}=90-A$

Similarly, we have $\angle{OBF}=90-C$ and $\angle{OCE}=90-B \implies \angle{DEC}=B$ and $\angle{DFB}=C$

Hence, quadrilaterals $AFDC$ and $AEDB$ are cyclic. $\angle{EDB}=\angle{FDB}=A$ . Let $DK'$ be the angle bisector of $\angle{EDF}$ , with $K'$ the point on the same side of $BC$ as $A$ such that $DEK'F$ is cyclic. Clearly $DK' \bot BC$ .

Also, since $K'$ is the midpoint of the arc $EF$ not including $D$ of $EDF$ 's circumcircle, $EK'=FK'$ . Since $EDFK'$ is cyclic, $\angle{EDK'}=\angle{EFK'}=90-A$ . Similarly, $\angle{FEK'}=90-A$ . Thus $K'$ is the unique point such that $\angle{EK'F}=2A$ and $EK'=FK'$ , so $K'$ is necessarily the circumcenter of $AEF$ . Thus, $K'=K$ and $DK \bot BC$ .

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yunxiu

571 posts

yunxiu

#3 h Apr 13, 2012, 5:01 AM • 11 Y

Y by jam10307, naw.ngs, Aspirant-to-IMO, Polynom_Efendi, myh2910, HappyMathEducation, egxa, Vladimir_Djurica, Adventure10, Mango247, and 1 other user

$\angle EDC = 90^\circ - \angle OCB = \angle A$ , similarly $\angle FDB = \angle A$ . So $\angle FKE = 2\angle A = 180^\circ - \angle FDE$ , hence $K,F,D,E$ is cyclic. So $\angle KDF = \angle KEF = 90^\circ - \angle A = 90^\circ - \angle FDB$ , we done.

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sunken rock

4384 posts

sunken rock

#4 h Apr 13, 2012, 1:47 PM • 3 Y

Y by Adventure10, Mango247, DensSv

Remark, open:

Locus of $K$ when $D$ moves randomly on $|BC|$ is a line segment.

Best regards,
sunken rock

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MBGO

315 posts

MBGO

#5 h Apr 13, 2012, 3:55 PM • 2 Y

Y by DownWithIsrael, Adventure10

sunken rock wrote:

Remark, open:

Locus of $K$ when $D$ moves randomly on $|BC|$ is a line segment.

Best regards,
sunken rock

Also your conjecture is not true...try to use Geogebra.

With Regards.

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yetti

2643 posts

yetti

#6 h Apr 14, 2012, 9:46 AM • 5 Y

Y by sunken rock, pentagon, Han1728, Adventure10, Mango247

yunxiu wrote:

Let $ABC$ be a triangle with circumcentre $O$ . The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$ . (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$ . Prove that the lines $DK$ and $BC$ are perpendicular.

Parallel to $BC$ through $F$ cuts $DE$ at $E'$ . $DE \perp CO$ is antiparallel to $AB$ WRT $\angle BCA$ $\Longrightarrow$ $\angle EE'F = \angle EDC = \angle CAB \equiv \angle EAF$ $\Longrightarrow$ $E' \in (K)$ , where $(K) \equiv \odot (AFE)$ .
Likewise, parallel to $BC$ through $E$ cuts $DF$ at $F'$ and $F' \in (K)$ . $EE'FF'$ is cyclic isosceles trapezoid inscribed in circle $(K)$ , with opposite sides $EE', FF'$ intersecting at $D$ .
By symmetry, $KD \perp (FE' \parallel EF' \parallel BC)$ . $\blacksquare$

sunken rock wrote:

Remark, open:
Locus of $K$ when $D$ moves randomly on $|BC|$ is a line segment.

When $D$ coincides with $B$ , then $F$ also coincides with $B$ . Let $(M) \equiv \odot (AFE)$ in this case. Since $MB \perp BC$ , circle $(M)$ is tangent to $BC$ at $B$ .
When $D$ coincides with $C$ , then $E$ also coincides with $C$ . Let $(N) \equiv \odot (AFE)$ in this case. Since $NC \perp BC$ , circle $(N)$ is tangent to $BC$ at $C$ .
Let $Z$ be intersection of circles $(M), (N)$ other than $A$ . Their radical axis $AZ$ cuts $BC$ at its midpoint $A'$ $\Longrightarrow$ $AZ \equiv AA'$ is A-median of $\triangle ABC$ .
From $(M), (N)$ tangent to $BC$ at $B, C,$ respectively $\Longrightarrow$ $\color[rgb]{0.75,0,0} \angle CZB = \pi - (\angle ZBC + \angle BCZ)= \pi - (\angle ZAB + \angle CAZ) = \pi - \angle CAB$ .

Let $P \in BC$ be foot of the A-symmedian of $\triangle ABC$ . Let $(L) \equiv \odot (AFE)$ in this case.
Then $DE, DF$ are tangents of $(L)$ at $E, F$ $\Longrightarrow$ $EF \parallel BC$ $\Longrightarrow$ intersection $Z' \equiv BE \cap CF$ is on A-median $AA'$ of $\triangle ABC$ .
Since $PE \perp CO, PF \perp BO$ are antiparallels to $AB, CA$ WRT $\angle BCA, \angle ABC$ $\Longrightarrow$ $ABPE$ , $AFPC$ are cyclic $\Longrightarrow$
$\color[rgb]{0.75,0,0} \angle CZ'B = \pi - (\angle Z'BC + \angle BCZ') = \pi - (\angle EBP + \angle PCF) =$ $\color[rgb]{0.75,0,0} \pi - (\angle CAP + \angle PAB) = \pi - \angle CAB$ .
It follows that points $Z' \equiv Z$ are identical, circles $(M), (N), (L)$ form a pencil intersecting at $A, Z$ and their centers $M, N, L$ collinear.

When $D \equiv P$ , points $E' \equiv E$ and points $F' \equiv F$ are identical.
When $D \in BC$ is arbitrary, isosceles $\triangle FDE', \triangle F'DE$ form centrally similar files with similarity centers $B, C,$ respectively $\Longrightarrow$
diagonal lines $BE', CF'$ of trapezoids $BDE'F, DCEF'$ are fixed $\Longrightarrow$ points $B, Z, E'$ are always collinear and points $C, Z, F'$ are always collinear.
From the cyclic isosceles trapezoid $EE'FF'$ $\Longrightarrow$ $\angle FZE = \angle F'ZE' = \angle CZB = \pi - \angle CAB = \pi - \angle EAF$ .
As a result, circles $(K) \equiv \odot(AFE)$ for arbitrary $D \in BC$ form a pencil intersecting at $A, Z$ and their centers $K$ lie on the line $MN$ .
If point $D$ is confined to the line segment $|BC|$ , then the circumcenters $K$ fill the line segment $|MN|$ . $\blacksquare$

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Rijul saini

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Rijul saini

#7 h Apr 18, 2012, 1:37 PM • 6 Y

Y by jam10307, CaptainCuong, shiningsunnyday, Polynom_Efendi, myh2910, Adventure10

yunxiu wrote:

Let $ABC$ be a triangle with circumcentre $O$ . The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$ . (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$ . Prove that the lines $DK$ and $BC$ are perpendicular.

Let circumradius of $\triangle AFE$ be $r$ . $\angle BDF =90^{\circ} - \angle OBD = \angle BAC$ , therefore, $D,F,A,C$ are concyclic, implying $BK^2 - r^2 = BD \cdot BC$

Similarly, $D,E,A,B$ concyclic, implying $CK^2 - r^2 = CD \cdot CB$ .

Thus, $BK^2 - CK^2 = BC ( BD - DC) = (BD+CD)(BD-DC) = BD^2 - DC^2$ , and we're done.

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subham1729

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subham1729

#8 h Aug 9, 2012, 6:35 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $\angle {FDE}=x$
So $\frac {FK}{KD}=\frac {KE}{KD}=\frac {Sin x}{Cos(B-C)}=\frac {Sin (2A+x)}{Cos(B-C)}$
So we get $A+x=\angle{KDB}=90^0$ . Done.

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pentagon

18 posts

pentagon

#9 h Apr 4, 2013, 10:38 AM • 2 Y

Y by Adventure10, Mango247

sunken rock wrote:

Remark, open:
Locus of $K$ when $D$ moves randomly on $|BC|$ is a line segment.

yetti wrote:

When $D$ coincides with $B$ , then $F$ also coincides with $B$ ... As a result, circles $(K) \equiv \odot(AFE)$ for arbitrary $D \in BC$ form a pencil intersecting at $A, Z$ and their centers $K$ lie on the line $MN$ .

The power of $B$ wrt $\odot (AFE)$ is $P(B)=BD \cdot BC$ . The power of $C$ wrt $\odot (AFE)$ is $P(C)=CD \cdot BC$ .
Let $M_{BC} \in BC$ be the midpoint of $BC$ . The power of $M_{BC}$ wrt $\odot (AFE)$ is $P(M_{BC})=\frac {2P(B)+2P(C)-BC^2}{4}=\frac{BC^2}{4}$ . This follows from the parallelogram law or the formula for the median. This means that $AM_{BC}$ is the radical axis of all circles $\odot (AFE)$ for an arbitrary position of $D \in BC$ . The centers $K$ of these circles lie on a perpendicular to this axis.

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Dukejukem

695 posts

Dukejukem

#10 h Oct 24, 2015, 12:55 AM • 2 Y

Y by Adventure10, Mango247

Since $DF \perp BO$ , it is well-known that $DF$ is antiparallel to $AC$ WRT $\angle ABC.$ Therefore, $A, C, D, F$ are concyclic. Similarly, $A, B, D, E$ are concyclic. Then after inversion with pole $A$ , we obtain the following

Equivalent Problem: Let $D$ be a variable point on the circumcircle of $\triangle ABC.$ Denote $E \equiv BD \cap AC, F \equiv CD \cap AB$ , and let $K$ be the reflection of $A$ in $EF.$ Prove that $\odot(ABC)$ and $\odot(ADK)$ are orthogonal.

Proof: Let the tangents to $\odot(ABC)$ at $A, D$ meet at $X$ . Note that the circle $\omega$ with center $X$ passing through $A, D$ is orthogonal to $\odot(ABC).$ Meanwhile, Pascal's Theorem applied to cyclic hexagon $AABDDC$ yields $X \in EF.$ But as $EF$ is the perpendicular bisector of $\overline{AK}$ , we obtain $XA = XK \implies K \in \omega.$ Hence, $\odot(ADK) \equiv \omega$ and the desired result follows. $\square$

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Dukejukem

695 posts

Dukejukem

#11 h Oct 24, 2015, 1:08 AM • 2 Y

Y by Adventure10, Mango247

sunken rock wrote:

Remark, open:

Locus of $K$ when $D$ moves randomly on $|BC|$ is a line segment.

Regarding the above post (#10), let $V$ be the projection of $A$ onto $EF$ (the midpoint of $\overline{AK}$ ) and let the tangents to $\odot(ABC)$ at $B, C$ meet at $T.$ By Pascal's Theorem applied to cyclic hexagon $ABBDCC$ , we obtain $T \in EF.$ Therefore, $\angle AVT = 90^{\circ}$ , and it follows that the locus of $T$ is the circle of diameter $\overline{AT}.$ Then the homothety $\mathbf{H}(A, 2)$ implies that the locus of $K$ is a circle passing through $A$ as well. Therefore, the locus of $K$ before inversion is a line not passing through $A.$

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HadjBrahim-Abderrahim

169 posts

HadjBrahim-Abderrahim

#13 h Mar 18, 2017, 5:38 PM • 3 Y

Y by azzam2912, Dr_Vex, Adventure10

yunxiu wrote:

Let $ABC$ be a triangle with circumcentre $O$ . The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$ . (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$ . Prove that the lines $DK$ and $BC$ are perpendicular.

Netherlands (Merlijn Staps)

Here is my solution. Let $X,Y$ be the intersection of $OB$ with $DF$ and $OC$ with $DE.$ Because the points $O$ $,$ $K$ are the circumcenters of the triangles $ABC$ $,$ $AEF,$ respectively, and the quadrilateral $OXDY$ is cyclic $( \angle OXD+\angle DYO=\frac{\pi}{2}+\frac{\pi}{2}=\pi ),$ we have $\angle EDF=\angle YDX=\pi-\angle BOC=\pi-2\angle BAC=\pi-\angle FKE.$ Therefore, the quadrilateral $KEDF$ is cyclic, because $KE=KF$ we deduce that $\angle EDK=\angle KDF.$ Because $OB=OC$ we can see that $\angle CDE=\angle CDY=\frac{\pi}{2}-\angle YCD=\frac{\pi}{2}-\angle XBD=\angle XDB=\angle FDB.$ Finally, we conclude that $\angle KDB=\angle KDF+\angle FDB=\angle EDK+\angle CDE=\angle CDK=\frac{\pi}{2},$ as claim.

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Muriatic

89 posts

Muriatic

#14 h Jun 15, 2017, 9:29 PM • 2 Y

Y by Adventure10, Mango247

This one can be done with the perpendicularity lemma.

The $CO\perp DE$ condition and $DF\perp BO$ are kind of a joke; they just mean that the point $E$ is on the circle $ABD$ , and $F$ is on the circle $ACD$ . But now we can let the points $M,N$ be the centres of the circles $ACD$ and $ABD$ , and the calculation from here is very quick: Since $NK$ is the perpendicular bisector of the segment $AE$ , we have the perpendicular lines $NK$ and $AC$ . Also we have the perpendicular $MK$ and $AB$ . Now we can just compute, where we used the signed lengths:

$\begin{aligned}BK^2-CK^2 &= (BK^2-AK^2) - (CK^2-AK^2) \\ &= (BM^2-AM^2) -(CN^2-AN^2) \\ &= P (B, ADC) -P (C, ABD)) \\ &= BD\cdot BC - CD\cdot CB \\ &= BD(BD+DC)-CD(CD+DB) \\ &= BD^2-CD^2 \end{aligned}$ But now it is obvious. (Here, $P$ denotes the power)

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Math1331Math

5317 posts

Math1331Math

#15 h Jun 15, 2017, 10:39 PM • 2 Y

Y by Adventure10, Mango247

sol

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uraharakisuke_hsgs

365 posts

uraharakisuke_hsgs

#16 h Nov 20, 2017, 5:15 PM • 2 Y

Y by Adventure10, Mango247

Nice result , here's my solution
Take $R$ on $BC$ such that $AR,AD$ are isogonal conjugate . $X$ on $BC$ such that $\triangle AXD \sim \triangle ARD$ , similar with $Y$ on $AC$ , then $AX.AC = AR.AD = AY.AB$ , then $XY \parallel BC$ .
We also have $\angle XDA = \angle C = \angle XFD$ so $XF.XA = XD^2$ , then $X$ lies on radical - axis of $(D,0)$ and $(K)$ . Similary with $Y$ , we have $XY \perp KD$ , which implies $BC \perp KD$

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andyxpandy99

365 posts

andyxpandy99

#50 h Jul 21, 2023, 1:14 PM

Y by

The key claim is that $KEDF$ is cyclic.

We prove this with angle chasing. Note that $\angle{OBC} = \angle{OCB} = 90-\angle{A}$ . It follows that $\angle{FDB} = \angle{EDC} = \angle{A}$ . This means that $\angle{EDF} = 180-\angle{FDB}-\angle{EDC} = 180-2\angle{A}$ . However, since $K$ is the circumcenter of $(AFE)$ , we know $\angle{FKE} = 2\angle{A}$ . Since $\angle{FKE} + \angle{EDF} = 2\angle{A} + 180-2\angle{A} = 180$ , we know that $KEDF$ is cyclic as desired.

This means that $\angle{KDF} = \angle{KEF} = 90-\angle{A}$ . Since $\angle{KDB} = \angle{KDF}+\angle{FDB} = 90-\angle{A} + \angle{A} = 90$ we have $KD \perp BC$ as desired.

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HamstPan38825

8857 posts

HamstPan38825

#51 h Aug 3, 2023, 8:18 PM

Y by

Perpendicularity lemma for the win

Note that $DFAC$ and $EDBA$ are cyclic as $\angle BFD = \angle C$ and $\angle BDF = A$ . Then $BK^2-CK^2=BF \cdot FA - BE \cdot BA = (BD+CD)(BD-CD)$ which finishes by perpendicularity lemma.

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pie854

243 posts

pie854

#52 h Aug 5, 2023, 11:56 AM

Y by

Math1331Math wrote:

sol

Huh, is the solution right? I want to know how come $|d|$ , $|e|$ , $|f|$ , $|k|$ are $1$ as they are not on the unit circle.

I am new to complex bashing but I tried to solve this question with it. I could not finish it as it got really ugly but if I had known that the absolute values are $1$ then a lot would have been easier.

This post has been edited 1 time. Last edited by pie854, Aug 5, 2023, 11:59 AM
Reason: Grammatical error

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ismayilzadei1387

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ismayilzadei1387

#53 h Nov 27, 2023, 7:58 AM • 1 Y

Y by FriIzi

$\angle OBC=\alpha=\angle OCB$
$\angle OAB=\beta=\angle OBA$
$\angle BFD=90-\beta$
$\angle BDF=90-\alpha$
$\angle DEC=\alpha+\beta$
$\angle KAO=x$
$\angle KFE=y$
$\angle AFK=\beta+x$
$\angle KAE=\angle KEA=90-x-y-\beta$
so $\angle EFD=90-x-y$
similarly $\angle FED=90+x-y$
so $y=\alpha$
which implies $KFDE$ cyclic
$\angle FDK=\alpha$
$\angle KDB=\angle FDB+\angle FDK=90$ indeed

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cj13609517288

1891 posts

cj13609517288

#54 h Dec 22, 2023, 7:51 PM

Y by

Note that
$\angle FDB=90^{\circ}-\angle OBC=\angle A.$ Similarly, $\angle EDC=\angle A$ as well. Therefore,
$\angle FKE=2\angle A=\angle FDB+\angle EDC=180^{\circ}-\angle FDE,$ so quadrilateral $KFDE$ is cyclic. Thus $\angle KDE=\angle KFE=\angle KEF=\angle KDF$ , and since $\angle FDB=\angle EDC$ , we have $\angle KDB=\angle KDC$ , so they are both $90^{\circ}$ , as desired. $\blacksquare$

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cursed_tangent1434

597 posts

cursed_tangent1434

#55 h Dec 26, 2023, 9:30 AM • 1 Y

Y by GeoKing

We start off with the following observation.

Claim : Quadrilaterals $ABDE$ and $AFDC$ are cyclic.

Proof : Note that,
$2\measuredangle EDC = 2\measuredangle OCD = 2\measuredangle OCB = \measuredangle COB = 2\measuredangle CAB$ Thus, $\measuredangle EDC = \measuredangle EAB$ implying that $ABDE$ is cyclic. With an entirely similar angle chase we obtain that $AFDC$ is also cyclic.

From this note that we have,
$\measuredangle EDC = \measuredangle CAB = \measuredangle BDF$ Now, we have the next claim.

Claim : $K$ lies on $(DEF)$ and is in fact the midpoint of arc $EF$ not containing $D$ .

Proof : Simply note that,
$\measuredangle EKF = 2\measuredangle EAF = 2\measuredangle CAB =\measuredangle EDF$ Thus, $KEDF$ is indeed cyclic and since we have $KE=KF$ this also implies that $K$ is the arc midpoint of $EF$ not containing $D$ .

Thus, $DK$ is the $\angle EDF$ -bisector and since we have $\measuredangle EDC = \measuredangle BDF$ as well, we can conclude that $KD \perp BC$ as desired.

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dolphinday

1323 posts

dolphinday

#56 h Jan 4, 2024, 1:00 PM

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Because $K$ is the circumcenter of $\triangle AFE$ , $\angle EFK = 2\angle A$ .
In addition, $\angle BOC = 2\angle A$ .
Then, because $OC = OB$ , $\angle OCB = \angle OBC = 90 - \angle A$ .
So $\angle CD = \angle BDF = \angle A \implies \angle EDF = 180 - 2\angle A \implies KFDE$ is cyclic.
Then because $KE = KF$ , $\angle EDK = \angle FDK = 90 - \angle A$ , so $\angle KDC = 90^{\circ}$ as desired.

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eibc

600 posts

eibc

#57 h Feb 22, 2024, 10:02 PM

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Because $CO$ and the $C$ -altitude are isogonal wrt $\angle C$ , we find that $\overline{AB}$ and $\overline{DE}$ are antiparallel in $\angle C$ , so $ABDE$ is cyclic. Similarly, $AFDC$ is cyclic. Now, we note that
$\measuredangle FDE = \measuredangle FDB + \measuredangle CDE = \measuredangle FDC + \measuredangle BDE = \measuredangle FAC + \measuredangle BAE = 2\measuredangle FAE = \measuredangle FKE,$ so $FKDE$ is cyclic. Thus, $K$ is the midpoint of $\widehat{FE}$ , so $KD$ bisects $\angle FDE$ . But from $\measuredangle BDF = \measuredangle CAF = \measuredangle EAB = \measuredangle EDC$ we see that $KD$ bisects straight $\angle BDC$ too, and $\overline{KD} \perp \overline{BC}$ .

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meth4life2020

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meth4life2020

#58 h Mar 13, 2024, 7:17 PM

Y by

Solved with epicbird08.

We claim that $DEKF$ is cyclic. Let $H$ be the orthocenter of $ABC$ . We see that $\angle BDF = 90 - \angle OBC = 90 - \angle HBA = \angle BAC$ due to $O$ and $H$ being isogonal conjugates with respect to $ABC$ . Similarly, $\angle CDE = \angle BAC$ . Therefore, $\angle EDF = 180 - 2\angle BAC$ . Additionally, because $(AEF)$ has center $K$ , $\angle EKF = 2\angle EAF = 2\angle BAC$ . Therefore, $\angle EKF + \angle EDF = 180$ , so $DEKF$ is cyclic.

Finally, because $EK = KF$ , we conclude that $K$ is the arc midpoint of arc $EF$ not containing $D$ , so $DK$ bisects $\angle EDF$ . However, then $\angle CDE + \angle EDK = \angle BAC + 90 - \angle BAC = 90$ , and we are done.

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dolphinday

1323 posts

dolphinday

#59 h Jun 8, 2024, 11:00 PM

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Re-solving for some reason
Let $CO \cap ED$ and $BO \cap DF$ be $P$ and $Q$ . Then $OPDQ$ is cyclic by opposite right angles. Then $\angle POQ = 2\angle A \implies \angle EDF = 180^\circ - 2\angle A$ . However $\angle EKF = 2\angle A$ so $DEKF$ is cyclic. Then $\angle KEF = \angle KFE = \angle KDF = \angle KDE = 90 - \angle A$ . However $\angle EDC = \angle A$ so $\angle KDC = 90^\circ$ as desired.

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HyperDunteR

41 posts

HyperDunteR

#60 h Jun 9, 2024, 12:27 AM

Y by

Since $OB = OC \implies \angle DBX = \angle DBX= \alpha \implies \angle XDB = \angle CDY = 90^{\circ} - \alpha \implies \angle XDY = \alpha$ .
To finish, we have that $2(90^{\circ} - \alpha) = 2 \angle FAB = \angle FKE \implies \angle FKE + \angle FDE = 180^{\circ} \implies F,K,E,D$ are concyclic $\implies \angle FDK = \angle FEK = \alpha = \angle KFE = \angle KDE \implies \angle KDB = \angle KDC = 90^{\circ} . \blacksquare$
Not much relevant

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dudade

139 posts

dudade

#62 h Jul 29, 2024, 12:35 AM

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Note that
$\angle DEF = 90 - \angle OCE = 90 - \angle OBF = \angle DFE.$ Thus, $\triangle DEF$ is isosceles. But, $K$ lies on the perpendicular bisector of $EF$ and $DK$ is an altitude of $\triangle DEF$ . So, $DK \perp BC$ , as desired.

This post has been edited 1 time. Last edited by dudade, Jul 29, 2024, 12:35 AM

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khanhnx

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khanhnx

#63 h Jul 29, 2024, 1:14 PM

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It's easy to see that $ACDF, ABDE$ are cyclic quadrilaterals. From this, we have $BK^2 - CK^2 = \mathcal{P}_{B / (AEF)} - \mathcal{P}_{C / (AEF)} = \overline{BF} \cdot \overline{BA} - \overline{CE} \cdot \overline{CA} = \overline{BD} \cdot \overline{BC} - \overline{CD} \cdot \overline{CB} = BD^2 - CD^2$ . Then $BC \perp DK$

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Mathandski

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Mathandski

#64 h Aug 26, 2024, 11:45 PM

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Subjective Rating (MOHs) $\text{[math]}$

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Avron

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Avron

#65 h Apr 15, 2025, 10:36 PM

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Notice that $\angle FKE+\angle FDE=2\angle A + 180-2\angle A=180$ so points $FDEK$ lie on a circle. For obvious reasons $\angle FDB=\angle EDC$ so it is sufficient to prove that $KD$ is the angle bisector of $\angle FDE$ , which follows from $FKED$ being cyclic and $FK=KE$

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