Interesting Properties related to Four 9-point Centers

by TelvCohl, Jul 17, 2016, 12:36 PM

Preliminaries

Lemma 1 (Gergonne-Euler theorem): Given a $ \triangle ABC $ and a point $ P. $ Let $ P_a, $ $ P_b, $ $ P_c $ be the reflection of $ P $ in $ BC, $ $ CA, $ $ AB, $ respectively. Then $$ \boxed{\frac{[\triangle P_aP_bP_c]}{[\triangle ABC]} = \left | \frac{\text{Power} (P, \odot (ABC))}{{R}^2} \right |} . $$
Proof : Let $ H $ be the orthocenter of $ \triangle ABC $ and let $ D, $ $ E $ be the second intersection of $ AP, $ $ AP_b $ with $ \odot (ABC), $ $ \odot (CHA), $ respectively. Clearly, $ A $ is the circumcenter of $ \triangle PP_bP_c, $ so $ \measuredangle AP_bP_c $ $ = $ $ 90^{\circ} $ $ - $ $ \measuredangle BAC $ $ = $ $ \measuredangle AEH $ $ \Longrightarrow $ $ P_bP_c $ $ \parallel $ $ EH, $ hence we get $ [\triangle HP_bP_c] $ $ = $ $ [\triangle EP_bP_c] $ $ = $ $ \tfrac{1}{2} $ $ \cdot $ $ P_bE $ $ \cdot $ $ P_bP_c $ $ \cdot $ $ \left| \cos \angle A \right | . $ Notice $ P_bE $ $ = $ $ PD $ (from symmetry) and $ P_bP_c $ $ = $ $ 2 $ $ \cdot $ $ AP $ $ \cdot $ $ \sin \angle A $ we get $$ [\triangle HP_bP_c] = \left| \frac{AP \cdot PD}{2} \right| \cdot \left| \sin 2\angle A \right| = \left| \frac{\text{Power}(P,\odot(ABC))}{2} \right| \cdot \left| \sin 2\angle A \right| . $$Similarly, we can prove $ [\triangle HP_cP_a] $ $ = $ $ \left| \frac{\text{Power}(P,\odot(ABC))}{2} \right| $ $ \cdot $ $ \left| \sin 2\angle B \right| $ and $ [\triangle HP_aP_b] $ $ = $ $ \left| \frac{\text{Power}(P,\odot(ABC))}{2} \right| $ $ \cdot $ $ \left| \sin 2\angle C\right | , $ so we conclude that $$ \frac{[\triangle P_aP_bP_c]}{[\triangle ABC]} = \frac{\sum [\triangle HP_bP_c]}{[\triangle ABC]} = \left| \frac{\text{Power}(P,\odot(ABC))}{2} \right | \cdot \frac{\sum \left| \sin 2\angle A \right|}{[\triangle ABC]} = \left|\frac{\text{Power} (P, \odot (ABC))}{{R}^2}\right|. \qquad \blacksquare $$
Lemma 2 (Gergonne-Ann theorem) : Given a $ \triangle ABC $ with points $ D $ $ \in $ $ BC, $ $ E $ $ \in $ $ CA, $ $ F $ $ \in $ $ AB. $ Let $ X, $ $ Y, $ $ Z $ be the points such that $ A $ $ \in $ $ YZ, $ $ B $ $ \in $ $ ZX, $ $ C $ $ \in $ $ XY $ and $ YZ $ $ \parallel $ $ EF, $ $ ZX $ $ \parallel $ $ FD, $ $ XY $ $ \parallel $ $ DE. $ Then $$ \boxed{{[\triangle ABC]}^2 = [\triangle DEF] \cdot [\triangle XYZ ]} .$$
Proof : Clearly, $ \triangle DEF $ and $ \triangle XYZ $ are homothetic, so $ DX, $ $ EY, $ $ FZ $ are concurrent at $ T. $ Let $ {\varsigma}^2 $ $ \cdot $ $ [\triangle DEF] $ $ = $ $ [\triangle XYZ] $ ($ \varsigma $ $ \in $ $ {\mathbb{R}}^{+} $). Since $$ \frac{[AETF]}{[\triangle TEF]} = \frac{\text{dist}(T,YZ)}{\text{dist}(T,EF)} = \varsigma, \frac{[BFTD]}{[\triangle TFD]} = \frac{\text{dist}(T,ZX)}{\text{dist}(T,FD)} = \varsigma, \frac{[CDTE]}{[\triangle TDE]} = \frac{\text{dist}(T,XY)}{\text{dist}(T,DE)} = \varsigma, $$so we conclude that $$ \frac{[\triangle ABC]}{[\triangle DEF]} = \varsigma = \sqrt{\frac{[\triangle XYZ]}{[\triangle DEF]}} \Longrightarrow {[\triangle ABC]}^2 = [\triangle DEF] \cdot [\triangle XYZ ]. \qquad \blacksquare $$
Lemma 3 : Given a $ \triangle ABC $ and a point $ P. $ Let $ Q $ be the image of $ P $ under the Inversion WRT $ \odot (ABC) $ and let $ \triangle P_aP_bP_c, $ $ \triangle Q_aQ_bQ_c $ be the pedal triangle of $ P, $ $ Q $ WRT $ \triangle ABC, $ respectively. Let $ M $ be the midpoint of $ PQ $ and let $ R $ be the isogonal conjugate of $ P $ WRT $ \triangle P_aP_bP_c. $ Then $ \triangle P_aP_bP_c $ $ \cup $ $ R $ $ \stackrel{-}{\sim} $ $ \triangle Q_aQ_bQ_c $ $ \cup $ $ M. $

Proof : Let $ B^*, $ $ C^* $ be the reflection of $ B, $ $ C $ in $ Q_c, $ $ Q_b, $ respectively. Since $ \triangle P_aP_bP_c $ and $ \triangle Q_aQ_bQ_c $ are inversely similar, so $ \measuredangle BPC $ $ = $ $ \measuredangle BAC $ $ + $ $ \measuredangle P_bP_aP_c $ $ = $ $ \measuredangle BAC $ $ + $ $ \measuredangle Q_cQ_aQ_b, $ hence notice that $ \measuredangle Q_cQ_aQ_b $ $ = $ $ \measuredangle (Q_cQ_a, BC) $ $ + $ $ \measuredangle (BC, Q_aQ_b) $ $ = $ $ \measuredangle Q_cQB $ $ + $ $ \measuredangle CQQ_b $ $ = $ $ \measuredangle B^*QQ_c $ $ + $ $ \measuredangle Q_bQC^* $ we get $ \measuredangle BPC $ $ = $ $ \measuredangle B^*QC^*. $ On the other hand, since $ \odot (ABC) $ is the A-apollonius circle of $ \triangle APQ, $ so $\tfrac{BP}{CP} $ $ = $ $ \tfrac{BQ}{CQ} $ $ = $ $ \tfrac{B^*Q}{C^*Q} $ $ \Longrightarrow $ $ \triangle BPC $ and $ \triangle B^*QC^* $ are directly similar $ \Longrightarrow $ $ \triangle B^*QC^* $ $ \stackrel{+}{\sim} $ $ \triangle Q_cMQ_b $ $ \stackrel{+}{\sim} $ $ \triangle BPC $ $ \stackrel{-}{\sim} $ $ \triangle P_cRP_b. $ Analogously, we can prove $ \triangle Q_aMQ_c $ $ \stackrel{-}{\sim} $ $ \triangle P_aRP_c $ and $ \triangle Q_bMQ_a $ $ \stackrel{-}{\sim} $ $ \triangle P_bRP_a, $ so we conclude that $$ \triangle P_aP_bP_c \cup R \stackrel{-}{\sim} \triangle Q_aQ_bQ_c \cup M. \qquad \blacksquare $$
Lemma 4 : Given a $ \triangle ABC $ and a point $ P. $ Let $ \triangle P_aP_bP_c $ be the pedal triangle of $ P $ WRT $ \triangle ABC $ and let $ O, $ $ T $ be the circumcenter of $ \triangle ABC, $ $ \triangle P_aP_bP_c, $ respectively. Let $ Q, $ $ R $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC, $ $ \triangle P_aP_bP_c, $ respectively and let $ U, $ $ V $ be the Poncelet point of $ ABCP, $ $ ABCQ, $ respectively. Then the Steiner line of $ U, $ $ V $ WRT $ \triangle P_aP_bP_c $ is parallel to $ OQ, $ $ TR, $ respectively.

Proof : Let $ \triangle Q_aQ_bQ_c $ be the pedal triangle of $ Q $ WRT $ \triangle ABC $ and let $ H, $ $ K $ be the projection of $ A $ on $ QQ_a, $ $ OQ, $ respectively. For any point $ I $ lying on the circumcircle of the triangle $ \mathbf{T}, $ let $ \mathcal{S}^{I}_{\mathbf{T} } $ be the Steiner line of $ I $ WRT $ \mathbf{T}. $ Since $ (U,K), $ $ (H, Q_a), $ are symmetry WRT the A-midline of $ \triangle ABC $ (well-known), so notice $ \mathcal{S}^{P_a}_{\triangle P_aP_bP_c} $ $ \perp $ $ P_bP_c $ we get $$ \measuredangle ( \mathcal{S}^{U}_{\triangle P_aP_bP_c}, \perp P_bP_c) = \measuredangle P_aQ_aU = \measuredangle KHA = \measuredangle KQA = \measuredangle (OQ, \perp P_bP_c) \Longrightarrow \mathcal{S}^{U}_{\triangle P_aP_bP_c} \parallel OQ. $$
Let $ \triangle P_AP_BP_C, $ $ \triangle R_AR_BR_C $ be the circumcevian triangle of $ P, $ $ R $ WRT $ \triangle ABC, $ $ \triangle P_aP_bP_c, $ respectively. It's obvious that $ P_bP_c $ $ \perp $ $ Q_aR_A, $ so notice $ \triangle P_AP_BP_C $ $ \cup $ $ O $ $ \cup $ $ P $ $ \stackrel{+}{\sim} $ $ \triangle P_aP_bP_c $ $ \cup $ $ T $ $ \cup $ $ R $ we conclude that $$ \measuredangle (\mathcal{S}^{V}_{\triangle P_aP_bP_c}, OP) = \measuredangle (\mathcal{S}^{V}_{\triangle P_aP_bP_c}, \mathcal{S}^{V}_{\triangle Q_aQ_bQ_c}) = \measuredangle (\mathcal{S}^{Q_a}_{\triangle P_aP_bP_c}, \mathcal{S}^{Q_a}_{\triangle Q_aQ_bQ_c}) = \measuredangle (P_aR, P_AP) = \measuredangle (TR, OP) \Longrightarrow \mathcal{S}^{V}_{\triangle P_aP_bP_c} \parallel TR. \qquad \blacksquare $$
Main result

Notation (all symbols in this section bear the same meaning)

Let $ P, $ $ Q $ be the isogonal conjugate of $ \triangle ABC. $ Let $ N, $ $ N_a, $ $ N_b, $ $ N_c $ be the 9-point center of $ \triangle ABC, $ $ \triangle PBC, $ $ \triangle PCA, $ $ \triangle PAB, $ respectively. Let $ \triangle Q_aQ_bQ_c $ be the pedal triangle of $ Q $ WRT $ \triangle ABC $ and let $ Q^* $ be the isogonal conjugate of $ Q $ WRT $ \triangle Q_aQ_bQ_c. $ Let $ M $ be the midpoint of $ PQ $ (center of the pedal circle of $ P, $ $ Q $ WRT $ \triangle ABC $).

Property 1 : $ \triangle N_aN_bN_c $ $ \cup $ $ N $ $ \cup $ $ M $ and $ \triangle Q_aQ_bQ_c $ $ \cup $ $ Q $ $ \cup $ $ Q^* $ are inversely similar.

Proof : Let $ M_{bc}, $ $ M_{ca}, $ $ M_{ab}, $ $ M_{ap}, $ $ M_{bp}, $ $ M_{cp} $ be the midpoint of $ BC, $ $ CA, $ $ AB, $ $ AP, $ $ BP, $ $ CP, $ respectively. Since $ \odot (N), $ $ \odot (M), $ $ \odot (N_a), $ $ \odot (N_b), $ $ \odot (N_c) $ are concurrent at the Poncelet point $ V $ of $ ABCP, $ so $$ \left\{\begin{array}{cc} \measuredangle NN_bN_c = \measuredangle M_{ca}VM_{ap} = \measuredangle M_{ca}M_{cp}M_{ap} = \measuredangle PAC = \measuredangle BAQ = \measuredangle Q_cQ_bQ \\\\ \measuredangle NN_cN_b = \measuredangle M_{ab}VM_{ap} = \measuredangle M_{ab}M_{bp}M_{ap} = \measuredangle PAB = \measuredangle CAQ = \measuredangle Q_bQ_cQ \end{array}\right\| \Longrightarrow \triangle NN_bN_c \stackrel{-}{\sim} \triangle QQ_bQ_c. $$Analogously, we can prove $ \triangle NN_cN_a $ $ \stackrel{-}{\sim} $ $ \triangle QQ_cQ_a $ and $ \triangle NN_aN_b $ $ \stackrel{-}{\sim} $ $ \triangle QQ_aQ_b, $ so $ \triangle N_aN_bN_c $ $ \cup $ $ N $ $ \stackrel{-}{\sim} $ $ \triangle Q_aQ_bQ_c $ $ \cup $ $ Q. $

Let $ \triangle P_aP_bP_c $ be the pedal triangle of $ P $ WRT $ \triangle ABC. $ Since $ \measuredangle MN_bN_c $ $ = $ $ \measuredangle P_bVM_{ap} $ $ = $ $ \measuredangle M_{cp}VM_{ca} $ ($\because $ $ P_bM_{ap} $ $ = $ $ M_{cp}M_{ca} $) $ = $ $ \measuredangle N_aN_bN, $ so $ N_bM, $ $ N_bN $ are isogonal conjugate WRT $ \angle N_cN_bN_a. $ Similarly, we can prove $ N_cM, $ $ N_cN $ are isogonal conjugate WRT $ \angle N_aN_cN_b, $ so $ M, $ $ N $ are isogonal conjugate WRT $ \triangle N_aN_bN_c $ $ \Longrightarrow $ $ \triangle N_aN_bN_c $ $ \cup $ $ N $ $ \cup $ $ M $ $ \stackrel{-}{\sim} $ $ \triangle Q_aQ_bQ_c $ $ \cup $ $ Q $ $ \cup $ $ Q^*. $ $ \qquad \blacksquare $

Property 2 : Let $ O $ be the circumcenter of $ \triangle ABC, $ then $ QQ^* $ $ \parallel $ $ OP $ and $ \frac{QQ^*}{2OP} =\frac{[\triangle Q_aQ_bQ_c]}{[\triangle ABC]}. $

Proof : Let $ \triangle Q_a^*Q_b^*Q_c^* $ be the pedal triangle of $ Q^* $ WRT $ \triangle Q_aQ_bQ_c $ and let $ J $ be its circumcenter (midpoint of $ QQ^* $). From $ Q_b^*Q_c^* $ $ \perp $ $ QQ_a, $ $ Q_c^*Q_a^* $ $ \perp $ $ QQ_b, $ $ Q_a^*Q_b^* $ $ \perp $ $ QQ_c $ we get $ Q_b^*Q_c^* $ $ \parallel $ $ BC, $ $ Q_c^*Q_a^* $ $ \parallel $ $ CA, $ $ Q_a^*Q_b^* $ $ \parallel $ $ AB, $ so combining $ Q^*Q_a^* $ $ \parallel $ $ PA, $ $ Q^*Q_b^* $ $ \parallel $ $ PB, $ $ Q^*Q_c^* $ $ \parallel $ $ PC $ we get $ \triangle Q_a^*Q_b^*Q_c^* $ $ \cup $ $ J $ $ \cup $ $ Q^* $ and $ \triangle ABC $ $ \cup $ $ O $ $ \cup $ $ P $ are homothetic $ \Longrightarrow $ $ QQ^* $ $ \parallel $ $ OP. $ From Lemma 2 $ \Longrightarrow $ $ [\triangle Q_a^*Q_b^*Q_c^*] $ $ \cdot $ $ [\triangle ABC ] $ $ = $ $ {[\triangle Q_aQ_bQ_c]}^2, $ so we conclude that $$ \frac{QQ^*}{2OP} = \frac{JQ^*}{OP} = \sqrt{\frac{[\triangle Q_a^*Q_b^*Q_c^*]}{[\triangle ABC]}} = \frac{[\triangle Q_aQ_bQ_c]}{[\triangle ABC]}. \qquad \blacksquare $$
Property 3 : Let $ R $ be the image of $ Q $ under the Inversion WRT $ \odot (O) $ and let $ \triangle R_aR_bR_c $ be the pedal triangle of $ R $ WRT $ \triangle ABC. $ Then $ \triangle N_aN_bN_c, $ $ \triangle R_aR_bR_c $ are homothetic with homothety center $ Q^*. $

Proof : Since the image of $ \odot (N), $ $ \odot (N_a), $ $ \odot (N_b), $ $ \odot (N_c) $ under the homothety $ (P,2) $ are concurrent at the antigonal conjugate $ S $ of $ P $ WRT $ \triangle ABC $ which is also the isogonal conjugate of $ R $ WRT $ \triangle ABC $ (well-known), so $ N_bN_c $ $ \parallel $ $ R_bR_c $ ($\perp $ $ AS $), $ N_cN_a $ $ \parallel $ $ R_cR_a $ ($\perp $ $ BS $), $ N_aN_b $ $ \parallel $ $ R_aR_b $ ($\perp $ $ CS $) $ \Longrightarrow $ $ \triangle N_aN_bN_c $ and $ \triangle R_aR_bR_c $ are homothetic. Let $ W, $ $ T $ be the midpoint of $ OP, $ $ QR, $ respectively. Obviously, $ \triangle N_aN_bN_c $ $ \cup $ $ W $ and $ \triangle R_aR_bR_c $ $ \cup $ $ R $ are homothetic. Since $ \tfrac{OQ}{OR} $ $ = $ $ \tfrac{{OQ}^2}{{R}^2}, $ so combining Lemma 1 and Property 2 we get $$ \frac{RQ}{RO} = \frac{ \left |{R}^2-{OQ}^2 \right |}{{R}^2} = \frac{4[\triangle Q_aQ_bQ_c]}{[\triangle ABC ]} = \frac{QQ^*}{OW} \Longrightarrow Q^* \in WR . $$
On the other hand, from $ \tfrac{RT}{WM} $ $ = $ $ \tfrac{RQ}{OQ} $ $ = $ $ \tfrac{RQ^*}{WQ^*} $ we know $ Q^* $ lies on $ MT, $ so combining $ \triangle N_aN_bN_c $ $ \cup $ $ M $ $ \stackrel{-}{\sim} $ $ \triangle Q_aQ_bQ_c $ $ \cup $ $ Q^* $ $ \stackrel{-}{\sim} $ $ \triangle R_aR_bR_c $ $ \cup $ $ T $ (Property 1 + Lemma 3) we conclude that $ Q^* $ $ \equiv $ $ WR $ $ \cap $ $ MT $ is the homothety center of $ \triangle N_aN_bN_c $ and $ \triangle R_aR_bR_c. $ $ \qquad \blacksquare $

Property 4 : $ T $ is the image of $ Q^* $ under the Inversion WRT $ \odot (M). $

Proof : Let $ R_M $ be the radius of $ \odot (M). $ Since $ \triangle Q_aQ_bQ_c $ and the circumcevian triangle of $ Q $ WRT $ \triangle ABC $ are directly similar with corresponding point $ (Q^*, Q), $ $ (M, O), $ so $ \tfrac{R}{OQ} $ $ = $ $ \tfrac{R_M}{MQ^*}, $ hence we conclude that $$ \frac{MQ^*}{MT} = \frac{OQ}{OR} = \frac{{OQ}^2}{{R}^2} = \frac{{MQ^*}^2}{{R_M}^2} \Longrightarrow MQ^* \cdot MT = {R_M}^2. \qquad \blacksquare $$
Property 5 : Let $ S_A, $ $ S_B, $ $ S_C $ be the center of $ \odot (SBC), $ $ \odot (SCA), $ $ \odot (SAB), $ respectively. Then the harmonic conjugate $ K $ of $ O $ WRT $ Q, $ $ R $ is the homothety center of of $ \triangle R_aR_bR_c $ $ \cup $ $ R $ and $ \triangle S_AS_BS_C $ $ \cup $ $ O. $

Proof : Since $ S_A, $ $ S_B, $ $ S_C $ is the reflection of $ P $ in $ N_a, $ $ N_b, $ $ N_c, $ respectively, so we conclude that $$ \frac{KR}{KO} = \frac{RR_a}{2WN_a} = \frac{RT}{2WM} = \frac{QR}{2QO} \Longrightarrow (O,K;Q,R) = -1. \qquad \blacksquare $$
Corollary 6 : Let $ P_A, $ $ P_B, $ $ P_C $ be the center of $ \odot (PBC), $ $ \odot (PCA), $ $ \odot (PAB), $ respectively. Then $ K $ is the homothety center of $ \triangle P_AP_BP_C $ $ \cup $ $ O $ and $ \triangle Q_aQ_bQ_c $ $ \cup $ $ Q. $

Property 7 : Let $ R^* $ be the isogonal conjugate of $ R $ WRT $ \triangle R_aR_bR_c. $ Then $ K $ lies on $ PQ^* $ and $ SR^*. $

Proof : Since $ Q^*(O,P; W,Q) $ $ = $ $ -1 $ $ = $ $ Q^*(O,K; R,Q), $ so $ K $ lies on $ PQ^*. $ Similarly, we can prove $ K $ $ \in $ $ SR^*. $ $ \qquad \blacksquare $

Property 8 : $ \triangle P_AP_BP_C $ and of $ \triangle S_AS_BS_C $ share the same orthocenter $ H^*. $

Proof : This is a particular case of Interesting Hexagon (Property 6). $ \qquad \blacksquare $

Property 9 : The circumcenter $ O_p $ of $ \triangle N_aN_bN_c $ is the complement (WRT $ \triangle ABC $) of the 9-point center $ N_P $ of the antipedal triangle $ \triangle X_PY_PZ_P $ of $ P $ WRT $ \triangle ABC. $

Proof : Let $ O_P^*, $ $ O_S^* $ be the circumcenter of $ \triangle P_AP_BP_C, $ $ \triangle S_AS_BS_C, $ respectively. Since $ S_A, $ $ S_B, $ $ S_C $ is the reflection of $ P_A, $ $ P_B, $ $ P_C $ in the midpoint of $ BC, $ $ CA, $ $ AB, $ respectively, so the centroid $ G $ of $ \triangle ABC $ is the midpoint of the centroid of $ \triangle P_AP_BP_C, $ $ \triangle S_AS_BS_C $ $ \Longrightarrow $ $ G $ is the centroid of $ \triangle H^*O_P^*O_S^*, $ hence the 9-point center $ N_P^* $ of $ \triangle P_AP_BP_C $ is the complement of $ O_S^* $ WRT $ \triangle ABC. $ Since $ O_p $ is the midpoint of $ PO_S^* $ and $ N_P $ is the reflection of $ P $ in $ N_P^*, $ so from Menelaus' theorem ($ \triangle PN_P^*O_S^* $ and $ \overline{GO_pN_P} $) we conclude that $ N_P $ is the anticomplement of $ O_p $ WRT $ \triangle ABC. $ $ \qquad \blacksquare $

Property 10 : $ O_p $ lies on $ MV. $

Proof : Let $ O_S $ be the circumcenter of the antipedal triangle of $ S $ WRT $ \triangle ABC. $ From Property 4 $ \Longrightarrow $ the reflection of $ S $ in $ Q $ is the image of $ S $ under the inversion WRT $ \odot (O_S), $ so $ O_S^* $ lies on $ QS $ $ \Longrightarrow $ the midpoint $ O_p, $ $ M, $ $ V $ of $ PO_S^*, $ $ PQ, $ $ PS $ are collinear. $ \qquad \blacksquare $

Corollary 11 : The circumcenter $ O_P $ of $ \triangle X_PY_PZ_P $ lies on $ PR. $ Furthermore, $ \odot (R,RP) $ $ \perp $ $ \odot (O_P), $ $ \odot (Q, QS) $ $ \perp $ $ \odot (O_S). $

Property 12 : Let $ R_P, $ $ R_S $ be the radius of $ \odot (O_P), $ $ \odot (O_S), $ respectively. Then $ R_P $ $ = $ $ SO_S, $ $ R_S $ $ = $ $ PO_P. $

Proof : From symmetry, it suffices to prove $ R_P $ $ = $ $ SO_S. $ Since the anticomplement $ U $ of $ V $ WRT $ \triangle ABC $ is the pole of the Simson line of $ \triangle ABC $ with direction $ \parallel $ $ OQ, $ so from Lemma 4 we know $ U $ is the orthopole of $ O_PP $ WRT $ \triangle X_PY_PZ_P, $ hence $ U $ $ \in $ $ \odot (N_P) $ $ \Longrightarrow $ $ V $ lies on the complement of $ \odot (N_P) $ WRT $ \triangle ABC. $ Combining Property 9 $ \Longrightarrow $ $ \odot (O_p, O_pV) $ is the complement of $ \odot (N_P) $ WRT $ \triangle ABC, $ so we conclude that $$ SO_S = 2 \cdot SO_S^* = 4 \cdot VO_p = 2 \cdot \left ( \text{radius of } \odot (N_P) \right) = R_P. \qquad \blacksquare $$
9-Point Center

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