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 $DK \bot BC$.

$\angle EDC = 90^\circ - \angle OCB = \angle A$, 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.

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

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.

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 $|MN|$. $\blacksquare$

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.

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.

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.

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. $\square$

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 $A.$

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.

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)

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 $BC \perp KD$

Remark - Let $DK$ meet circumcircle of triangle $AFE$ at $I$ such that $I$ lies inside triangle $DFE$. Then $I$ is the incenter of triangle $DFE$.

My proof for the original problem. $\angle FDB=\angle A$ so $\measuredangle FKE = \measuredangle FDE$, so $FKED$ is cyclic. Now $KE=KF$, so $K$ is the midpoint of arc $EF$. Now $K$ is on the line perpendicular to $BC$, since $DK$ bisects the angle $EDF$. sa2001 wrote: Remark - Let $DK$ meet circumcircle of triangle $AFE$ at $I$ such that $I$ lies inside triangle $DFE$. Then $I$ is the incenter of triangle $DFE$. Apply fact 5.

Another solution: Clearly $A,C,D,F$ and $A,B,D,E$ are circles. Now, by easy angle chase $BE\cap CF$ lies on the circumcircle of $(AFE)$. Now $D$ is Miquel point of $AF(BE\cap CF)E$ so instantly done...

Let $X = FD \cap OB$ and $Y = ED \cap OC$. Note that $OXDY$ is cyclic because $\angle{OXD} + \angle{OYD} = 90+90 = 180$. Thus, $\angle{FDE} = 180-2\angle{A}$. Further note that $\angle{FKE} = 2\angle{A}$ implying that $KFDE$ is also cyclic. We have that $\angle{KDE} = \angle{KFE} = 90-\angle{A}$ Because $\angle{OCB}= 90-\angle{A}$, we have $\angle{EDC} = \angle{A}$. Thus, $\angle{KDC} = 90$, finishing the proof.

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.

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.

Math1331Math wrote:

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.

$\angle OBC=\alpha=\angle OCB$ $\angle OAB=\beta=\angle OBA$ $\angle BFD=90-\beta$ $\angle BDF=90-\alpha$ $\angle DEC=\alpha+\beta$ $\angle KAO=x$ $\angle KFE=y$ $\angle AFK=\beta+x$ $\angle KAE=\angle KEA=90-x-y-\beta$ 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

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. $\blacksquare$

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.

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.

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 $\overline{KD} \perp \overline{BC}$.

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.

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.

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 $\implies \angle FDK = \angle FEK = \alpha = \angle KFE = \angle KDE \implies \angle KDB = \angle KDC = 90^{\circ} . \blacksquare$

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.

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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