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Assisted perpendicular chasing
sarjinius   4
N 2 hours ago by X.Allaberdiyev
Source: Philippine Mathematical Olympiad 2025 P7
In acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D$ be an arbitrary point on the circumcircle of triangle $ABC$ such that $D$ does not lie on line $OB$ and that line $OD$ is not parallel to line $BC$. Let $E$ be the point on the circumcircle of triangle $ABC$ such that $DE$ is perpendicular to $BC$, and let $F$ be the point on line $AC$ such that $FA = FE$. Let $P$ and $R$ be the points on the circumcircle of triangle $ABC$ such that $PE$ is a diameter, and $BH$ and $DR$ are parallel. Let $M$ be the midpoint of $DH$.
(a) Show that $AP$ and $BR$ are perpendicular.
(b) Show that $FM$ and $BM$ are perpendicular.
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sarjinius
Mar 9, 2025
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The intersection of two lines lies on circumcircle
LeVietAn   1
N Jun 20, 2023 by Paramizo_Dicrominique
Source: OWN
Let $ABC$ be a triangle with $I$ be the incenter of its. Let $\omega$ be a circle passing through $I$ and intersects $AI$, $BI$, $CI$ and circumcircle of triangle $IBC$ at $D$, $E$, $F$ and $G$, respectively ($I$ different from $D$, $E$, $F$, $G$). The line $DG$ intersects $CA$, $AB$ at $M$, $N$, respectively. The perpendicular bisector of $AD$ intersects $DE$, $DF$ at $K$, $L$, respectively. Prove that the intersection of $KM$ and $LN$ is on circumcircle of triangle $ABC$.
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LeVietAn
Dec 16, 2016
Paramizo_Dicrominique
Jun 20, 2023
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The intersection of two lines lies on circumcircle
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Source: OWN
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LeVietAn
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LeVietAn
#1 h  Dec 16, 2016, 4:52 PM • 3 Y
Y by baopbc, anhtaitran, Adventure10
Let $ABC$ be a triangle with $I$ be the incenter of its. Let $\omega$ be a circle passing through $I$ and intersects $AI$, $BI$, $CI$ and circumcircle of triangle $IBC$ at $D$, $E$, $F$ and $G$, respectively ($I$ different from $D$, $E$, $F$, $G$). The line $DG$ intersects $CA$, $AB$ at $M$, $N$, respectively. The perpendicular bisector of $AD$ intersects $DE$, $DF$ at $K$, $L$, respectively. Prove that the intersection of $KM$ and $LN$ is on circumcircle of triangle $ABC$.
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Paramizo_Dicrominique
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Paramizo_Dicrominique
#2 h  Jun 20, 2023, 3:27 PM
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$\textbf{Solution:}$
We will solve for the case when $\triangle{ABC}$ is acute, the obtuse case is the same.
Let $(AKL)$ cut $(ABC)$ again at $W$. We will prove $NL$ passes through $W$ and similarly $MK$ passes through $W$.
Let $DK$ cut $AB$ at $U$, $DL$ cut $AC$ at $V$.
We have $\angle{UDL}= \angle{EDF} = 180^{\circ} - \angle{EIF} = 180^{\circ} - \angle{BIC} = 90^{\circ} - \frac{1}{2}\angle{BAC} = 90^{\circ} - \angle{UAD}$
Since $L$ lie on the perpendicular bisector of $AD$ so $L$ is the center of $(AUD)$. Similarly $K$ is the center of $(AVD)$.
We have $\angle{DKV} = 90^{\circ} - \angle{DAV} = 90^{\circ} - \angle{DAU} = \angle{ULD}$ $\implies A,K,U,L,V,W$ lie on a circle.
Let $AI$ cut $(AUV)$ again at $H$ so $H$ is the midpoint of arc $UV$ of $(AUV)$.
Let $WH$ cut $UV$ at $P$. We will prove $\overline{D,G,P}.$
Let $DG$ cut $UV$ at $P'$ so $\frac{P'U}{P'V} = \frac{sin(P'DU)}{sin(P'DV)} \cdot \frac{DU}{DV} = \frac{sin(GDE)}{sin(GDF)} \cdot \frac{DU}{DV}=\frac{sin(GFE)}{sin(GEF)} \cdot \frac{DU}{DV} = \frac{GE}{GF} \cdot \frac{DU}{DV} = \frac{BE}{CF} \cdot \frac{DU}{DV} $ (Since $\triangle{GEB} \overset{+}{\sim} \triangle{GFC}$)
Back working on the ratio of $P$,we have $\frac{PU}{PV} = \frac{WU}{WV} = \frac{BU}{CV}$ (Since $\triangle{WUB} \overset{+}{\sim} \triangle{WVC}$)
To get the claim $\overline{D,G,P}$ we will show $P \equiv P'$ or $\frac{PU}{PV}=\frac{P'U}{P'V}$
It is equivalent to $\frac{BU}{CV} = \frac{BE}{CF} \cdot \frac{DU}{DV} \Longleftrightarrow \frac{BU}{BE} \cdot \frac{CF}{CV} = \frac{DU}{DV} \Longleftrightarrow \frac{sin(BEU)}{sin(BUE)} \cdot \frac{sin(CVF)}{sin(CFV)} = \frac{sin(DVU)}{sin(DUV)} \Longleftrightarrow \frac{sin(DEI)}{sin(BUE)} \cdot \frac{sin(CVF)}{sin(DFI)} = \frac{sin(DVU)}{sin(DUV)} \Longleftrightarrow \frac{sin(CVF)}{sin(BUE)} = \frac{sin(DVU)}{sin(DUV)}$
But since $\angle{UDV} = \angle{EDF} = 180^{\circ} - \angle{EIF} = 180^{\circ} - \angle{BIC} = 90^{\circ} - \frac{1}{2}\angle{BAC}$ and $D$ lie on the bisector of $\angle{UAV}$ so $D$ is the $A-$excenter of $\triangle{AUV}$ so $\angle{BUE}=\angle{DUV},\angle{CVF} = \angle{DVU}$ which implies the equation above so that $\overline{D,G,P}.$
Let $S$ be the second intersection of $(WDV)$ with $AD$.
Let $DW$ cut $(AUV)$ again at $W'$ ,observe $H$ is the midpoint of arc $UV$ which is the center of $(UDV)$ so $\angle{ADG} = \angle{ADP} =\angle{HDP}=\angle{W'WH}=\angle{W'AH}$ but since $KL$ is the perpendicular bisector of $AD$ so $AW'$ cut $DP$ at a point $J$ lie on $KL$.
We will prove $\triangle{ASV} \overset{+}{\sim} \triangle{AND}$.
Since $\angle{AND} = 180^{\circ} - \angle{BND} = 180^{\circ} - \angle{BAD} - \angle{ADN} = 180^{\circ} - \angle{CAD} - \angle{JAD} = 180^{\circ} - \angle{JAV} = 180^{\circ} - \angle{W'AV} = 180^{\circ} - \angle{W'WV} = 180^{\circ} - \angle{DWV} = 180^{\circ} - \angle{DSV} = \angle{ASV}$
$\implies \angle{AND} = \angle{ASV}$ also since $\angle{NAD} = \angle{SAV}$ so we get the claim.
By spiral homothety so if $Q$ is the intersection of $SV,ND$ so $Q$ lie on $(ASN)$ and $(AVD)$.
We will use a well known lemma
$\textbf{Lemma:}$ Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$. Suppose diagonals $AC,BD$ cut at $P$. Prove that $(APB),(DPC),(OAD)$ has a common point.
Using the lemma for cyclic quadrilateral $QAVD$ with circumcenter $K$ so $W$ lie on $(QASN)$.
We have $\angle{LAD} = \angle{LDA} = \angle{VDA} = \angle{VQA} \implies$ $(QASNW)$ is tangent to $AL$ so that $\angle{WNA} = 180^{\circ} - \angle{WQA} = 180^{\circ} - \angle{WAL} = 180^{\circ} - \angle{WAL} = \angle{WUL}$ and since $L$ is the midpoint of arc $AU$ of $(AUW)$ so it is well known that $\overline{W,N,L}$.
Similarly we can prove $\overline{M,W,K}$ $\implies MK$ cut $NL$ at $W$ lie on $(ABC)$.
$Q.E.D.$
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This post has been edited 1 time. Last edited by Paramizo_Dicrominique, Jun 20, 2023, 3:28 PM
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