Main result

In this section, I'll give a brief introduction to Miquel triangle, Isogonal center and Gergonne-Steiner point of a quadrangle. All symbols in this section bear the same meaning.

Notation

Given a quadrangle $P_1P_2P_3P_4.$ Let $M_X, M_Y, M_Z$ be the Miquel point of the complete quadrilateral $\{ P_1P_2, P_1P_3, P_2P_4, P_3P_4 \},$ $\{ P_1P_2, P_2P_3, P_1P_4, P_3P_4 \},$ $\{ P_1P_3, P_2P_3, P_1P_4, P_2P_4 \},$ respectively and let $X, Y, Z$ be the intersection of $\{ P_1P_4, P_2P_3 \},$ $\{ P_1P_3, P_2P_4 \},$ $\{ P_1P_2, P_3P_4 \},$ respectively. Consider the transformation $\Psi_X$ defined by inversion with center $M_X,$ power $M_XM_Y \cdot M_XM_Z$ followed by reflection in the angle bisector of $\angle M_YM_XM_Z.$ For two points $J,K,$ we say $\{ J,K \} \in \mathbb{X}$ if $J, K$ is the image of $K, J$ under $\Psi_X.$ Similarly, we can define $\Psi_Y,\Psi_Z, \mathbb{Y}, \mathbb{Z}.$ Note that $\{ J,K \} \in \mathbb{X}$ $\Longleftrightarrow$ $\triangle M_XM_YJ \stackrel{+}{\sim} \triangle M_XKM_Z.$

Property 1 : $\{ P_1, P_4 \}, \{ P_2, P_3 \}, \{ Y,Z \} \in \mathbb{X},$ $\{ P_1, P_3 \}, \{ P_2, P_4 \}, \{ X,Z \} \in \mathbb{Y},$ $\{ P_1, P_2 \}, \{ P_3, P_4 \}, \{ X,Y \} \in \mathbb{Z}.$

Proof : Since $M_X$ is the Miquel point of the complete quadrilateral $\{ P_1P_2, P_1P_3, P_2P_4, P_3P_4 \},$ so we get $M_XP_1 \cdot M_XP_4 = M_XP_2 \cdot M_XP_3 = M_XY \cdot M_XZ = \varsigma$ and $\{ M_XP_1, M_XP_4 \}, \{ M_XP_2, M_XP_3 \}$ are isogonal conjugate WRT $\angle YM_XZ.$ Consider the transformation $\Phi_X$ define by inversion with center $M_X,$ power $\varsigma$ followed by reflection in the angle bisector of $\angle YM_XZ,$ then $\odot (P_2P_3Z) \mapsto \odot (P_2P_3Y) \odot (P_1P_4Z) \mapsto \odot (P_1P_4Y)$ under $\Phi_X,$ so we conclude that $M_Z$ is the image of $M_Y$ under $\Phi_X$ $\Longrightarrow$ $\Phi_X = \Psi_X.$ i.e. $\{ P_1, P_4 \}, \{ P_2, P_3 \}, \{ Y,Z \} \in \mathbb{X}.$ $\qquad \blacksquare$

Property 2 : The isogonal conjugate $T_i$ of $P_i$ WRT $\triangle M_XM_YM_Z$ is the isogonal conjugate of the complement of $P_i$ WRT $\triangle P_jP_kP_l$ WRT $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4\}.$

Proof : It suffices to prove the case when $i=4.$ Let $M_{ij}$ be the midpoint of $P_iP_j$ where $i, j \in \{ 1,2,3,4\}$ and let $P^*_4$ be the reflection of $P_4$ in $M_{23}.$ From $\triangle M_XP_1P_2 \stackrel{+}{\sim} \triangle M_XP_3P_4$ we get $\frac{P_2P_4^*}{M_XP_3} = \frac{P_3P_4}{M_XP_3} = \frac{P_1P_2}{M_XP_1}$ and $\measuredangle (P_1P_2, P_2P_4^*) = \measuredangle (P_1P_2, P_3P_4) = \measuredangle (M_XP_1, M_XP_3),$ so $\triangle P_1P_2P_4^* \stackrel{+}{\sim} \triangle P_1M_XP_3$ $\Longrightarrow$ $\{ P_1M_X, P_1P_4^* \}$ are isogonal conjugate WRT $\angle P_1.$ i.e. $P_1M_X$ passes through the isogonal conjugate of the complement of $P_4$ WRT $\triangle P_1P_2P_3$ WRT $\triangle P_1P_2P_3.$ Similarly, $P_2M_Y, P_3M_Z$ pass through the isogonal conjugate of the complement of $P_4$ WRT $\triangle P_1P_2P_3$ WRT $\triangle P_1P_2P_3.$ On the other hand, from Property 1 we know $\{ P_1, P_4 \} \in \mathbb{X},$ so $T_4$ lies on $P_1M_X.$ Similarly, $T_4$ lies on $P_2M_Y,$ $P_3M_Z,$ so we conclude that $T_4$ is the isogonal conjugate of the complement of $P_4$ WRT $\triangle P_1P_2P_3$ WRT $\triangle P_1P_2P_3.$ $\qquad \blacksquare$

Property 3 : $\{ T_1,T_4 \}, \{ T_2, T_3 \} \in \mathbb{X}, \{ T_1, T_3 \}, \{ T_2,T_4 \} \in \mathbb{Y}, \{ T_1, T_2 \}, \{ T_3,T_4 \} \in \mathbb{Z}$ and $P_1T_1 \parallel P_2T_2 \parallel P_3T_3 \parallel P_4T_4.$

Proof : From Property 1 we know $\{ P_1, P_4 \} \in \mathbb{X},$ so $\triangle M_XM_YP_1 \stackrel{+}{\sim} \triangle M_XP_4M_Z$ $\Longrightarrow$ $\measuredangle T_4P_1M_Y = \measuredangle M_XM_ZP_4 = \measuredangle T_4M_ZM_Y,$ so $T_4 \in \odot (P_1M_YM_Z).$ Similarly, we can prove $T_1, T_2, T_3$ is the second intersection of $P_4M_X, P_4M_Y, P_4M_Z$ with $\odot (P_4M_YM_Z),$ $\odot (P_4M_ZM_X),$ $\odot (P_4M_XM_Y),$ respectively, so $\triangle M_XM_ZT_2 \stackrel{+}{\sim} \triangle M_XT_3M_Y$ and $\{ T_1,T_4 \}, \{ T_2, T_3 \} \in \mathbb{X}.$ From $\{ P_1,P_4 \}, \{ T_1, T_4 \} \in \mathbb{X}$ $\Longrightarrow$ $\frac{M_XP_1}{M_XT_4} = \frac{M_XT_1}{M_XP_4},$ so $P_1T_1$ and $P_4T_4$ are parallel. Analogously, we can prove $\{ T_1, T_3 \}, \{ T_2,T_4 \} \in \mathbb{Y}, \{ T_1, T_2 \}, \{ T_3,T_4 \} \in \mathbb{Z}$ and $P_2T_2 \parallel P_4T_4,$ $P_3T_3 \parallel P_4T_4,$ so we conclude that $P_1T_1 \parallel P_2T_2 \parallel P_3T_3 \parallel P_4T_4.$ $\qquad \blacksquare$

Property 4 : $\odot (P_1P_4M_X), \odot (P_2P_3M_X), \odot (P_1P_3M_Y), \odot (P_2P_4M_Y), \odot (P_1P_2M_Z), \odot (P_3P_4M_Z)$ are concurrent at $I$ and $\{ I,X \} \in \mathbb{X},$ $\{ I,Y \} \in \mathbb{Y},$ $\{ I,Z \} \in \mathbb{Z}.$

Proof : Let $I$ be the second intersection of $\odot (P_1P_2M_Z), \odot (P_1P_3M_Y).$ Notice $\odot (P_1P_2M_Z), \odot (P_1P_3M_Y)$ is the image of $\odot (P_3P_4M_Y), \odot (P_2P_4M_Z)$ under $\Psi_X,$ respectively, so $\{ I,X \} \in \mathbb{X}.$ Combining $\{ P_2, P_3 \} \in \mathbb{X}$ we get $\triangle M_XP_2I \stackrel{+}{\sim} \triangle M_XXP_3$ $\Longrightarrow$ $\measuredangle P_2IM_X = \measuredangle P_2P_3M_X$ $\Longrightarrow$ $I \in \odot (P_2P_3M_X).$ Similarly, by $\{ P_1, P_4 \} \in \mathbb{X}$ we get $\triangle M_XP_1I \stackrel{+}{\sim} \triangle M_XXP_4,$ so $\measuredangle P_1IM_X = \measuredangle P_1P_4M_X$ and $I \in \odot (P_1P_4M_X).$ Analogously, we can prove $I$ lies on $\odot (P_2P_4M_Y),$ $\odot (P_3P_4M_Z),$ so we conclude that $\odot (P_1P_4M_X), \odot (P_2P_3M_X), \odot (P_1P_3M_Y), \odot (P_2P_4M_Y), \odot (P_1P_2M_Z), \odot (P_3P_4M_Z)$ are concurrent at $I$ and $\{ I,X \} \in \mathbb{X}.$ $\qquad \blacksquare$

Corollary 5 : $IP_i$ is the common tangent of $\odot (P_iM_XX), \odot (P_iM_YY), \odot (P_iM_ZZ)$ where $i \in \{ 1,2,3,4 \}.$

Proof : It suffices to prove the case when $i=4.$ From $I \in \odot (P_1P_4M_X)$ and $\triangle M_XP_1I \stackrel{+}{\sim} \triangle M_XXP_4$ $\Longrightarrow$ $\measuredangle IP_4M_X = \measuredangle IP_1M_X = \measuredangle P_4XM_X,$ so $IP_4$ is tangent to $\odot (P_4M_XX).$ Similarly, we can prove $IP_4$ is tangent to $\odot (P_4M_YY), \odot (P_4M_ZZ),$ so $IP_4$ is the common tangent of $\odot (P_4M_XX), \odot (P_4M_YY), \odot (P_4M_ZZ).$ $\qquad \blacksquare$

Corollary 6 : $M_XX, M_YY, M_ZZ$ are concurrent at $G.$

Proof : From $\{ I,X \} \in \mathbb{X}$ $\Longrightarrow$ $\{ M_XI, M_XX \}$ are isogonal conjugate WRT $\angle M_YM_XM_Z,$ so the isogonal conjugate $G$ of $I$ WRT $\triangle M_XM_YM_Z$ lies on $M_XX.$ Similarly, we can prove $G \in M_YY,$ $G \in M_ZZ.$ $\qquad \blacksquare$

Remark : $G$ is called the Gergonne-Steiner point and $I$ is called the Isogonal center of the quadrangle $P_1P_2P_3P_4.$

Property 7 : $G$ lies on $\odot (XM_YM_Z), \odot (YM_ZM_X), \odot (ZM_XM_Y).$

Proof : From the proof of Property 3 $\Longrightarrow$ $T_4$ lies on $\odot (P_2M_ZM_X),$ $\odot (P_3M_XM_Y),$ so $\measuredangle M_YM_XM_Z = \measuredangle M_YM_XT_4 + \measuredangle T_4M_XM_Z = \measuredangle M_YP_3T_4 + \measuredangle T_4P_2M_Z.$ On the other hand, since $I$ lies on $\odot (P_1P_2M_Z)$ and $\odot (P_1P_3M_Y),$ so $\measuredangle M_ZIM_Y = \measuredangle M_ZIP_1 + \measuredangle P_1IM_Y = \measuredangle M_ZP_2P_1 + \measuredangle P_1P_3M_Y,$ hence keep in mind that $G,I$ are isogonal conjugate WRT $\triangle M_XM_YM_Z$ we conclude that $\measuredangle M_YGM_Z = \measuredangle M_YM_XM_Z + \measuredangle M_ZIM_Y = \measuredangle M_YP_3T_4 + \measuredangle T_4P_2M_Z + \measuredangle M_ZP_2P_1 + \measuredangle P_1P_3M_Y = \measuredangle P_1P_3M_Z + \measuredangle M_YP_2P_1 = \measuredangle P_1XM_Z + \measuredangle M_YXP_1 = \measuredangle M_YXM_Z.$ i.e. $G$ lies on $\odot (XM_YM_Z).$ $\qquad \blacksquare$

Property 8 : $G$ is the complement of the antigonal conjugate of $P_i$ WRT $\triangle P_jP_kP_l$ WRT $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4\}.$

Proof : It suffices to prove the case when $i=4.$ Since $M_Y$ is the center of the spiral similarity of $\overline{P_1P_4M_{14}} \mapsto \overline{P_2P_3M_{23}},$ so $\measuredangle M_{14}M_YM_{23} = \measuredangle P_1M_YP_2 = \measuredangle M_{14}XM_{23}$ $\Longrightarrow$ $M_Y$ lies on $\odot (XM_{14}M_{23}).$ Analogously, we can prove $M_Z \in \odot (XM_{14}M_{23}),$ so $M_{14}$ and $M_{23}$ lie on $\odot (XM_YM_Z).$ Similarly, we can prove $M_{13}, M_{24} \in \odot (YM_ZM_X),$ $M_{12}, M_{34} \in \odot (ZM_XM_Y),$ so $\measuredangle M_{12}GM_{13} = \measuredangle M_{12}GM_X + \measuredangle M_XGM_{13} = \measuredangle M_{12}M_{34}M_X + \measuredangle M_XM_{24}M_{13} = \measuredangle P_2P_4M_X + \measuredangle M_XP_4P_3 = \measuredangle P_2P_4P_3,$ hence the anticomplement $G^*_4$ of $G$ WRT $\triangle P_1P_2P_3$ lies on the reflection of $\odot (P_2P_3P_4)$ in $P_2P_3.$ Analogously, we can prove $G^*_4$ lies on the reflection $\odot (P_1P_3P_4), \odot (P_1P_2P_4)$ in $P_1P_3,$ $P_1P_2,$ respectively, so we conclude that $G^*_4$ is the antigonal conjugate of $P_4$ WRT $\triangle P_1P_2P_3.$ i.e. $G$ is the complement of the antigonal conjugate of $P_4$ WRT $\triangle P_1P_2P_3$ WRT $\triangle P_1P_2P_3.$ $\qquad \blacksquare$

Corollary 9 : $G$ lies on $\odot (M_{12}M_{13}M_{14}), \odot (M_{12}M_{23}M_{24}), \odot (M_{13}M_{23}M_{34}), \odot (M_{14}M_{24}M_{34}).$

Property 10 : Let $S_i$ be the Miquel associate of $P_i$ WRT $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4\},$ then $I$ lies on $\odot (P_2S_1Z),$ $\odot (P_3S_1Y),$ $\odot (P_4S_1X),$ $\odot (P_1S_2Z),$ $\odot (P_3S_2X),$ $\odot (P_4S_2Y),$ $\odot (P_1S_3Y),$ $\odot (P_2S_3X),$ $\odot (P_4S_3Z),$ $\odot (P_1S_4X),$ $\odot (P_2S_4Y),$ $\odot (P_3S_4Z).$

Proof : It suffices to prove $I$ lies on $\odot (P_1S_4X).$ Consider an inversion with center $I$ and denote the image of $\Box$ as ${\Box}^{\mathbb{I}},$ then by Property 4 $\Longrightarrow$ $\triangle M_X^{\mathbb{I}}M_Y^{\mathbb{I}}M_Z^{\mathbb{I}}$ is the diagonal triangle of the quadrangle $P_1^{\mathbb{I}}P_2^{\mathbb{I}}P_3^{\mathbb{I}}P_4^{\mathbb{I}}.$ Since $X^{\mathbb{I}}$ lies on $\odot (P_2^{\mathbb{I}}P_4^{\mathbb{I}}M_Z^{\mathbb{I}}),$ $\odot (P_3^{\mathbb{I}}P_4^{\mathbb{I}}M_Y^{\mathbb{I}}),$ so $X^{\mathbb{I}}$ is the Miquel point of the complete quadrilateral $\{ P_1^{\mathbb{I}}P_2^{\mathbb{I}}, P_1^{\mathbb{I}}P_3^{\mathbb{I}}, P_2^{\mathbb{I}}P_4^{\mathbb{I}}, P_3^{\mathbb{I}}P_4^{\mathbb{I}} \}.$ Analogously, $Y^{\mathbb{I}}, Z^{\mathbb{I}}$ is the Miquel point of the complete quadrilateral $\{ P_1^{\mathbb{I}}P_2^{\mathbb{I}}, P_2^{\mathbb{I}}P_3^{\mathbb{I}}, P_1^{\mathbb{I}}P_4^{\mathbb{I}}, P_3^{\mathbb{I}}P_4^{\mathbb{I}} \},$ $\{ P_1^{\mathbb{I}}P_3^{\mathbb{I}}, P_2^{\mathbb{I}}P_3^{\mathbb{I}}, P_1^{\mathbb{I}}P_4^{\mathbb{I}}, P_2^{\mathbb{I}}P_4^{\mathbb{I}} \},$ respectively. Since $S_4^{\mathbb{I}}$ lies on $\odot (P_1^{\mathbb{I}}Y^{\mathbb{I}}Z^{\mathbb{I}}),$ $\odot (P_2^{\mathbb{I}}Z^{\mathbb{I}}X^{\mathbb{I}}),$ $\odot (P_3^{\mathbb{I}}X^{\mathbb{I}}Y^{\mathbb{I}}),$ so $S_4^{\mathbb{I}} \in P_1^{\mathbb{I}}X^{\mathbb{I}},$ by the proof of Property 3, hence we conclude that $I$ lies on $\odot (P_1S_4X).$ $\qquad \blacksquare$

Property 11 : $I$ is the image of the isogonal conjugate of $P_i$ WRT $\triangle P_jP_kP_l$ under the inversion WRT $\odot (P_jP_kP_l)$ where $\{ i,j,k,l \} = \{ 1,2,3,4\}.$

Proof : It suffices to prove the case when $i=4.$ Let $I^*$ be the image of the isogonal conjugate of $P_4$ WRT $\triangle P_1P_2P_3$ under the inversion WRT $\odot (P_1P_2P_3).$ Since $I$ lies on $\odot (P_2P_3M_X),$ so $\measuredangle P_2IP_3 = \measuredangle P_2M_XP_3 = \measuredangle P_2M_XP_4 + \measuredangle P_4M_XP_3 = \measuredangle P_2ZP_3 + \measuredangle P_2YP_3 = \measuredangle P_2P_1P_3 + \measuredangle P_2P_4P_3 = \measuredangle P_2I^*P_3.$ Similarly, we can prove $\measuredangle P_3IP_1 = \measuredangle P_3I^*P_1,$ $\measuredangle P_1IP_2 = \measuredangle P_1I^*P_2,$ so we conclude that $I \equiv I^*.$ i.e. $I$ is the image of the isogonal conjugate of $P_4$ WRT $\triangle P_1P_2P_3$ under the inversion WRT $\odot (P_1P_2P_3).$ $\qquad \blacksquare$

Property 12 : $S_i$ is the cevian quotient $(P_i / T_i)$ of $P_i, T_i$ WRT $\triangle P_jP_kP_l$ where $\{ i,j,k,l \} = \{ 1,2,3,4\}.$

Proof : It suffices to prove the case when $i=4.$ Let $\widehat{P}_4$ be the isotomic conjugate of $P_4$ WRT $\triangle P_1P_2P_3,$ $\triangle \widehat{X}_4\widehat{Y}_4\widehat{Z}_4$ be the cevian triangle of $\widehat{P}_4$ WRT $\triangle P_1P_2P_3,$ and $R_1, R_2, R_3$ be the second intersection of $\odot (P_1P_2P_3)$ with $\odot (P_1YZ),$ $\odot (P_2ZX),$ $\odot (P_3XY),$ respectively. Since $R_1$ is the center of the spiral similarity of $P_2Z \mapsto P_3Y,$ so we get $\frac{R_1P_2}{R_1P_3} = \frac{P_2Z}{P_3Y} = \frac{P_1\widehat{Z}_4}{P_1\widehat{Y}_4}$ $\Longrightarrow$ $\triangle R_1P_2P_3 \stackrel{+}{\sim} \triangle P_1\widehat{Z}_4\widehat{Y}_4,$ hence $P_1R_1$ is tangent to $\odot (P_1\widehat{Y}_4\widehat{Z}_4)$ at $P_1.$ Similarly, we can prove $P_2R_2, P_3R_3$ is tangent to $\odot (P_2\widehat{Z}_4\widehat{X}_4), \odot (P_3\widehat{X}_4\widehat{Y}_4)$ at $P_2,$ $P_3,$ respectively. Let $\triangle T_4^1T_4^2T_4^3$ be the triangle formed by $P_1R_1, P_2R_2, P_3R_3$ and $V_4^1, V_4^2, V_4^3$ be the isogonal conjugate of $T_4^1, T_4^2, T_4^3$ WRT $\triangle P_1P_2P_3,$ respectively. Note that $\triangle \widehat{X}_4\widehat{Y}_4\widehat{Z}_4$ and $\triangle V_4^1V_4^2V_4^3$ are homothetic, so $\triangle V_4^1V_4^2V_4^3$ is the anticevian triangle of the isotomcomplement of $\widehat{P}_4$ WRT $\triangle P_1P_2P_3$ WRT $\triangle P_1P_2P_3,$ hence by Property 2 we conclude that $\triangle T_4^1T_4^2T_4^3$ is the anticevian triangle of $T_4$ WRT $\triangle P_1P_2P_3$ $\Longrightarrow$ $S_4,$ the perspector of $\triangle XYZ$ and $\triangle T_4^1T_4^2T_4^3,$ is the cevian quotient $(P_4 / T_4 )$ of $P_4, T_4$ WRT $\triangle P_1P_2P_3.$ $\qquad \blacksquare$

Property 13 : $I$ lies on $S_iT_i$ where $i \in \{ 1,2,3,4 \}.$

Proof : It suffices to prove the case when $i=4.$ Let $\triangle Q_1Q_2Q_3$ be the circumcevian triangle of $T_4$ WRT $\triangle P_1P_2P_3,$ then $P_4^1Q_1 \cdot P_4^1P_1 = P_4^1X \cdot P_4^1S_4,$ so $Q_1$ lies on $\odot (P_1S_4X).$ Similarly, we can prove $Q_2 \in \odot (P_2S_4Y),$ $Q_3 \in \odot (P_3S_4Z),$ so from $P_1T_4 \cdot Q_1T_4 = P_2T_4 \cdot Q_2T_4 =P_3T_4 \cdot Q_3T_4$ we conclude that $T_4$ lies on the radical axis of $\odot (P_1S_4X), \odot (P_2S_4Y), \odot (P_3S_4Z).$ i.e. $I$ lies on $S_4T_4.$ $\qquad \blacksquare$

Property 14 : $T_1$ lies on $XS_4, YS_3, ZS_2.$ $T_2$ lies on $XS_3, YS_4, ZS_1.$ $T_3$ lies on $XS_2, YS_1, ZS_4.$ $T_4$ lies on $XS_1, YS_2, ZS_3.$

Proof : It suffices to prove $T_4$ lies on $XS_1.$ Since $\odot (IP_2Z), \odot (IP_3Y)$ is the image of $\odot (P_3XY), \odot (P_2ZX)$ under $\Psi_X,$ respectively, so $\{ S_1, S_4 \} \in \mathbb{X}.$ Combining Property 13 we get $M_X, S_4, T_4, X$ are concyclic. Similarly, we can prove $M_X, S_1, T_1, X$ are concyclic, so note that $\{ T_1, T_4 \} \in \mathbb{X},$ by Property 3, we conclude that $\measuredangle M_XXT_4 = \measuredangle M_XS_4T_4 = \measuredangle M_XT_1S_1 = \measuredangle M_XXS_1$ $\Longrightarrow$ $T_4\in XS_1.$ $\qquad \blacksquare$

Application

Problem 1 (Generalization of Musselman's theorem) : Given a $\triangle ABC$ with isogonal conjugate $P,Q.$ Let $D, E, F$ be the second intersection of $AQ, BQ, CQ$ with $\odot (BQC),$ $\odot (CQA),$ $\odot (AQB),$ respectively. Then the circles $\odot (APD), \odot (BPE), \odot (CPF)$ are coaxial.

Proof : Simple angle chasing yields $\triangle ABF \stackrel{+}{\sim} \triangle AEC,$ $\triangle ABD \stackrel{+}{\sim} \triangle APC,$ so we get $\frac{AE}{AP} = \frac{AD}{AF}$ and $\measuredangle PAE = \measuredangle FAD$ $\Longrightarrow$ $\triangle AEP \stackrel{+}{\sim} \triangle ADF,$ hence $A$ is the center of the spiral similarity of $EP \mapsto DF$ $\Longrightarrow$ $A$ is the Miquel point of the complete quadrilateral $\{ DE, DF, PE, PF\}.$ Similarly, we can prove $B,C$ is the Miquel point of the complete quadrilateral $\{ EF, ED, PF, PD\},$ $\{ FD, FE, PD, PE\},$ respectively, so by Property 4 we conclude that $\odot (APD), \odot (BPE), \odot (CPF)$ are coaxial. $\qquad \blacksquare$

Remark : Musselman's theorem is the case when $P, Q$ is the circumcenter, orthocenter of $\triangle ABC,$ respectively.

Problem 2 (2012 IMO shortlist G8) : Let $\triangle ABC$ be a triangle with circumcircle $\omega$ and $\ell$ be a line without common points with $\omega.$ Denote by $P$ the foot of the perpendicular from the center of $\omega$ to $\ell.$ The side-lines $BC, CA, AB$ intersect $\ell$ at the points $X, Y, Z$ different from $P.$ Prove that $\odot (APX), \odot (BPY), \odot (CPZ)$ have a common point different from $P$ or are mutually tangent at $P.$

Proof :

Lemma : Given a $\triangle ABC$ with isogonal conjugate $P,Q.$ Let $\triangle DEF$ be the cevian triangle of $P$ WRT $\triangle ABC$ and $X\in BC$ be the point s.t. $AX$ is tangent to $\odot (AEF)$ at $A.$ Then $X$ lies on the polar of $Q$ WRT $\odot (ABC).$

Proof : Let $BP, CP$ cuts $\odot (AEF)$ again at $P_B, P_C,$ respectively and $\triangle Q_aQ_bQ_c$ be the circumcevian triangle of $Q$ WRT $\triangle ABC.$ From Pascal's theorem for $EP_BP_CFAA$ we get $BC, P_BP_C$ and the tangent of $\odot (AEF)$ at $A$ are concurrent, so $X$ lies on $P_BP_C.$ Since $\measuredangle BAQ_b = \measuredangle BEC = \measuredangle P_BAX,$ so combining $\measuredangle ABQ_b = \measuredangle P_BBX$ we get $Q_b$ is the isogonal conjugate of $P_B$ WRT $\triangle AXB$ $\Longrightarrow$ $XQ_b$ is the isogonal conjugate of $P_BP_C$ WRT $\angle AXD.$ Similarly, we can prove $XQ_c$ is the isogonal conjugate of $P_BP_C$ WRT $\angle AXD,$ so we conclude that $X \in Q_bQ_c$ $\Longrightarrow$ $X$ lies on the polar of $Q$ WRT $\odot (ABC).$ $\qquad \square$

Back to the main proof :

Let $R$ be the isogonal conjugate of the image of $P$ under the inversion WRT $\omega$ WRT $\triangle ABC$ and $T$ be the Miquel associate of $R$ WRT $\triangle ABC,$ it suffices to prove $T$ lies on $\odot (APX), \odot (BPY), \odot (CPZ).$ Let $\triangle DEF$ be the cevian triangle of $R$ WRT $\triangle ABC,$ then by Lemma $\Longrightarrow$ $X$ lies on the tangent of $\odot (AEF)$ at $A,$ so $\measuredangle TDX = \measuredangle TFA = \measuredangle TAX$ $\Longrightarrow$ $A, D, T, X$ are concyclic. Note that $P$ is the Isogonal center of the quadrangle $ABCR,$ so by Property 10 we conclude that $A, D, P, T, X$ are concyclic. i.e. $T \in \odot (APX).$ $\qquad \blacksquare$

Remark : The condition $\ell \cap \omega = \varnothing$ is redundant.

Problem 3 : Given a $\triangle ABC$ with isogonal conjugate $P,Q.$ Let $\triangle DEF$ be the antipedal triangle of $P$ WRT $\triangle ABC$ and $R, S$ be the image of $Q, P$ under the inversion WRT $\odot (ABC),$ $\odot (DEF),$ respectively. Then $P, R, S$ are collinear and $R$ is the midpoint of $PS.$

Proof : Let $\triangle XYZ$ be the cevian triangle of $P$ WRT $\triangle ABC$ and $U, V, W$ be the Miquel point of the complete quadrilateral $\{ AB, AC, PB, PC\},$ $\{ BC, BA, PC, PA\},$ $\{ CA, CB, PA, PB\},$ respectively. Consider an inversion with center $P$ and denote the image of $\Box$ as ${\Box}^*,$ then $X^*, Y^*, Z^*$ is the second intersection of $A^*P^*, B^*P^*, C^*P^*$ with $\odot (B^*PC^*),$ $\odot (C^*PA^*),$ $\odot (A^*PB^*),$ respectively and $U^*, V^*, W^*$ is the intersection of $\{ B^*Z^*, C^*Y^* \},$ $\{ C^*X^*, A^*Z^* \},$ $\{ A^*Y^*, B^*X^* \},$ respectively. Furthermore, by Property 4 we get $R^*$ lies on $A^*U^*, B^*V^*, C^*W^*.$

Note that $\{ U^*, X^* \}, \{ V^*, Y^* \}, \{ W^*, Z^* \}$ are isogonal conjugate WRT $\triangle A^*B^*C^*,$ so $R^*$ is the isogonal conjugate of $P$ WRT $\triangle A^*B^*C^*.$ On the other hand, $\triangle D^*E^*F^*$ is the pedal triangle of $P$ WRT $\triangle A^*B^*C^*,$ so $S^*$ is the center of $\odot (D^*E^*F^*)$ $\Longrightarrow$ $S^*$ is the midpoint of $PR^*,$ hence we conclude that $P, R, S$ are collinear and $R$ is the midpoint of $PS.$ $\qquad \blacksquare$

Remark : See Interesting Properties related to Four 9-point Centers (Property 4) for another proof to this problem.