Properties related to the orthocenters of BPC, CPA, APB

by henderson, Jun 4, 2016, 8:54 PM

$\bf\color{red}Properties\ related \ to \ the \ orthocenters \ of \ \triangle BPC, \triangle CPA\ and \ \triangle APB$ $=============================================================================================$
In this topic, I'll prove some properties related to the orthocenter $J_a,$ $J_b,$ $J_c$ of $\triangle BPC,$ $\triangle CPA,$ $\triangle APB,$ respectively (where $P$ is an arbitrary point). Note that all symbols in this post bear the same meaning (except $(\spadesuit)$ ).

Property 1 : Let $Q$ be the isogonal conjugate of $P$ WRT $\triangle ABC$ and let $\triangle Q_aQ_bQ_c$ be its pedal triangle WRT $\triangle ABC.$ Then $\triangle Q_aQ_bQ_c$ is inscribed in $\triangle J_aJ_bJ_c$ (i.e. $Q_a,$ $Q_b,$ $Q_c$ lies on $J_bJ_c,$ $J_cJ_a,$ $J_aJ_b,$ respectively.).

Proof : Let $\triangle P_AP_BP_C$ be the antipedal triangle of $P$ WRT $\triangle ABC.$ Since $A,$ $C,$ $P,$ $P_B$ lie on a circle with diameter $PP_B,$ so $\measuredangle PP_BP_C$ $=$ $\measuredangle PCA$ $=$ $\measuredangle BCQ.$ Similarly, we can prove $\measuredangle PP_CP_B$ $=$ $\measuredangle CBQ,$ so we get $\triangle PP_BP_C$ $\cup$ $A$ $\stackrel{-}{\sim}$ $\triangle QCB$ $\cup$ $Q_a.$ Combining $BJ_c$ $\stackrel{\parallel}{=}$ $P_CA,$ $CJ_b$ $\stackrel{\parallel}{=}$ $P_BA$ $\Longrightarrow$ $\tfrac{BQ_a}{CQ_a}$ $=$ $\tfrac{BJ_c}{CJ_b}.$ i.e. $Q_a$ $\in$ $J_bJ_c.$ $\blacksquare$
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Property 2 : Let $H_P$ be the orthocenter of $\triangle P_AP_BP_C$ and let $M$ be the midpoint of $PH_P.$ Then the anticomplement $T$ of $M$ WRT $\triangle ABC$ is the orthocenter of $\triangle AJ_bJ_c,$ $\triangle BJ_cJ_a,$ $\triangle CJ_aJ_b.$

Proof : Let $G$ be the centroid of $\triangle ABC.$ Clearly, $G$ is the centroid of $\triangle PTH_P,$ so the midpoint of $TH_P$ is the complement of $P$ WRT $\triangle ABC.$ Notice $J_a$ is the reflection of $P_A$ in the midpoint of $BC$ we get $J_aT$ $\parallel$ $P_AH_P$ $\Longrightarrow$ $J_aT$ $\parallel$ $AP.$ Analogously, we can prove $J_bT$ $\parallel$ $BP$ and $J_cT$ $\parallel$ $CP,$ so $T$ is the orthocenter of $\triangle AJ_bJ_c,$ $\triangle BJ_cJ_a,$ $\triangle CJ_aJ_b.$ $\blacksquare$
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Property 3 : Let $K_a,$ $K_b,$ $K_c$ be the orthocenter of $\triangle BQC,$ $\triangle CQA,$ $\triangle AQB,$ respectively. Then the A-altitude of $\triangle ABC,$ $J_bJ_c,$ $K_bK_c$ are concurrent (similar for $(J_cJ_a , K_cK_a)$ and $(J_aJ_b , K_aK_b)$ ).

Proof : Let $AP$ cuts $\odot (BPC)$ at $R_a$ and let $T_a$ be the projection of $R_a$ on $BC$ (define $R_b,$ $R_c,$ $T_b,$ $T_c$ similarly). Let $H_A,$ $H_a$ be the orthocenter of $\triangle P_ABC,$ $\triangle R_aBC,$ respectively. From $\measuredangle R_aBC$ $=$ $\measuredangle R_aPC$ $=$ $\measuredangle P_CP_BP_A,$ $\measuredangle R_aCB$ $=$ $\measuredangle R_aPB$ $=$ $\measuredangle P_BP_CP_A$ $\Longrightarrow$ $\triangle R_aBC$ $\cup$ $H_a$ $\stackrel{-}{\sim}$ $\triangle P_AP_BP_C$ $\cup$ $H_P,$ so notice $P_AH_A$ $\stackrel{\parallel}{=}$ $R_aH_a$ we get $\frac{P_AH_A}{P_AH_P} = \frac{\text{dist}(R_a,BC)}{\text{dist}(P_A,P_BP_C)} = \frac{R_aT_a}{R_aA},$ hence $\triangle P_AH_AH_P,$ $\triangle R_aT_aA$ are homothetic $\Longrightarrow$ $AT_a$ is parallel to the Steiner line $H_AH_P$ of the complete quadrilateral $\mathcal{Q}$ formed by $\triangle P_AP_BP_C$ and $BC.$

Let $\triangle P_aP_bP_c$ be the pedal triangle of $P$ WRT $\triangle ABC.$ Let $A^*,$ $B^*,$ $C^*$ be the isotomic conjugate of $A,$ $B,$ $C$ WRT $(P_B,P_C),$ $(P_C,P_A),$ $(P_A,P_B),$ respectively. From $P_AB^*$ $\stackrel{\parallel}{=}$ $BP_C$ $\stackrel{\parallel}{=}$ $AJ_c,$ $P_AC^*$ $\stackrel{\parallel}{=}$ $CP_B$ $\stackrel{\parallel}{=}$ $AJ_b$ we get $\triangle AJ_bJ_c$ and $\triangle P_AC^*B^*$ are congruent and homothetic, so $J_bJ_c$ $\stackrel{\parallel}{=}$ $B^*C^*$ $\Longrightarrow$ $J_bJ_c$ is parallel to the Newton line of $\mathcal{Q},$ hence $AT_a$ $\perp$ $J_bJ_c.$

Let the perpendicular from $P_a$ to $J_bJ_c$ cuts the A-altitude of $\triangle ABC$ at $S_a$ and let $U$ $\equiv$ $PP_a$ $\cap$ $AT_a.$ Since $\frac{UP}{AP} = \frac{T_aR_a}{AR_a} = \frac{\text{dist}(R_a,BC)}{\text{dist}(P_A,P_BP_C)} = \frac{BC}{P_BP_C} = \frac{QQ_a}{AP},$ so $AS_a$ $=$ $PP_a$ $+$ $QQ_a.$ Similarly, we can prove $Q_aS_a$ $\perp$ $K_bK_c,$ so notice the intersection of $J_bJ_c,$ $K_bK_c$ is the orthocenter of $\triangle P_aQ_aS_a$ we conclude that the intersection of $J_bJ_c$ and $K_bK_c$ lies on the A-altitude of $\triangle ABC.$ $\blacksquare$

From the proof above we get the following corollaries :

Corollary 3.1 : $\triangle T_aT_bT_c$ is the cevian triangle of $T$ WRT $\triangle ABC.$

Corollary 3.2 : $\triangle J_aJ_bJ_c$ and $\triangle A^*B^*C^*$ are congruent and homothetic.
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Property 4 : $J_bK_c$ $\parallel$ $J_cK_b,$ $J_cK_a$ $\parallel$ $J_aK_c,$ $J_aK_b$ $\parallel$ $J_bK_a.$

Proof : Let $AP$ cuts $\odot (ABC)$ again at $A_P$ and $A_P^*$ be the projection of $A_P$ on $BC.$ Note that $\tfrac{AP}{PA_P}$ $=$ $\tfrac{\text{dist}(Q,BC)}{\text{dist}(A_P,BC)}$ (See here (Lemma at post #4) or here (Lemma)), so we get $\frac{PU}{A_PA_P^*}=\frac{\text{dist}(Q,BC)}{\text{dist}(A_P,BC)}=\frac{AP}{PA_P} \Longrightarrow PA_P^*\parallel AU \parallel P_aS_a.$ Analogously, if $AQ$ cuts $\odot (ABC)$ again at $A_Q$ and $A_Q^*$ is the projection of $A_Q$ on $BC,$ then $QA_Q^*$ $\parallel$ $Q_aS_a.$ Let $V_a$ (on the A-altitude of $\triangle ABC$ ) be the intersection of $J_bJ_c$ and $K_bK_c.$ Let $\infty_{\varsigma}$ be the point at infinity with direction $\varsigma .$ Since $A(J_b,J_c; V_a,\infty_{ \parallel J_bJ_c}) = P(C,B;\infty_{ \parallel BC},A_P^*) = Q(B,C;\infty_{ \parallel BC},A_Q^*) = A(K_c,K_b;V_a,\infty_{ \parallel K_bK_c}),$ so we conclude that $\tfrac{J_bV_a}{J_cV_a}$ $=$ $\tfrac{K_cV_a}{K_bV_a}$ $\Longrightarrow$ $J_bK_c$ $\parallel$ $J_cK_b.$ $\blacksquare$
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Property 5 : $\triangle J_aJ_bJ_c$ and $\triangle K_aK_bK_c$ are cyclologic.

Proof : Let $K$ be the second intersection of $\odot (J_bK_cK_a)$ and $\odot (J_cK_aK_b).$ Since $\measuredangle K_bKK_c = \measuredangle K_bKK_a + \measuredangle K_aKK_c = \measuredangle K_bJ_cK_a + \measuredangle K_aJ_bK_c = \measuredangle J_bK_cJ_a + \measuredangle J_aK_bJ_c = \measuredangle K_bJ_aK_c,$ so $K$ lies on $\odot (J_aK_bK_c).$ i.e. $\odot (J_aK_bK_c),$ $\odot (J_bK_cK_a),$ $\odot (J_cK_aK_b)$ are concurrent at $K.$

Analogously, we can prove $\odot (K_aJ_bJ_c),$ $\odot (K_bJ_cJ_a),$ $\odot (K_cJ_aJ_b)$ are concurrent at $J.$ $\blacksquare$
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Property 6 : Let $X_a$ $\equiv$ $P_bP_c$ $\cap$ $Q_bQ_c,$ $X_b$ $\equiv$ $P_cP_a$ $\cap$ $Q_cQ_a,$ $X_c$ $\equiv$ $P_aP_b$ $\cap$ $Q_aQ_b.$ Then $(J_a,K_a),$ $(J_b,K_b),$ $(J_c,K_c)$ lie on $X_bX_c,$ $X_cX_a,$ $X_aX_b,$ respectively.

Proof : It is well-known that $AX_a$ $\parallel$ $BX_b$ $\parallel$ $CX_c$ ( $\perp PQ$ ), so from Pappus's theorem for $\overline{Q_a B C},$ $\overline{\infty_{\perp PQ} \infty_{\perp BP} \infty_{\perp CP}}$ $\Longrightarrow$ $J_a,$ $X_b,$ $X_c$ are collinear. Similarly, we can prove $K_a$ $\in$ $X_bX_c,$ so $J_a,$ $K_a,$ $X_b,$ $X_c$ are collinear. $\blacksquare$
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$(\spadesuit)$ Before stating property 7, we recall following two lemmas.

Lemma 1 : Given a hexagon $A_1B_2C_1A_2B_1C_2$ s.t. $B_2C_1$ $\parallel$ $B_1C_2,$ $C_2A_1$ $\parallel$ $C_1A_2,$ $A_2B_1$ $\parallel$ $A_1B_2.$ Let $X_1$ be the intersection of $\odot (A_2B_1C_1),$ $\odot (A_1B_2C_1),$ $\odot (A_1B_1C_2)$ and $X_2$ be the intersection of $\odot (A_1B_2C_2),$ $\odot (A_2B_1C_2),$ $\odot (A_2B_2C_1).$ Then $A_1X_1 \parallel A_2X_2, B_1X_1 \parallel B_2X_2, C_1X_1 \parallel C_2X_2.$ Proof : Let $D_1$ $\in$ $\odot (A_2B_1C_1)$ be the point s.t. $\measuredangle D_1B_1C_1$ $=$ $\measuredangle A_2B_1C_2,$ $\measuredangle D_1C_1B_1$ $=$ $\measuredangle A_2C_1B_2$ and $D_2$ $\in$ $\odot (A_1B_2C_2)$ be the point s.t. $\measuredangle D_2B_2C_2$ $=$ $\measuredangle A_1B_2C_1,$ $\measuredangle D_2C_2B_2$ $=$ $\measuredangle A_1C_2B_1.$ From $\measuredangle D_1X_1B_1$ $=$ $\measuredangle D_1C_1B_1$ $=$ $\measuredangle A_2C_1B_2$ $=$ $\measuredangle A_1C_2B_1$ $=$ $\measuredangle A_1X_1B_1$ $\Longrightarrow$ $A_1 ,$ $D_1 ,$ $X_1$ are collinear. Similarly, we can prove $A_2,$ $D_2,$ $X_2$ are collinear.

Since $\triangle B_1C_1D_1$ $\stackrel{+}{\sim}$ $\triangle B_2C_2D_2,$ so notice $\measuredangle A_2B_1D_1$ $=$ $\measuredangle B_2C_1B_1,$ $\measuredangle A_1B_2D_2$ $=$ $\measuredangle C_1B_2C_2$ we get $\frac{A_1D_2}{A_2D_1} = \frac{B_2C_2}{B_1C_1} \cdot \frac{ \sin \angle C_1B_2C_2}{\sin \angle B_2C_1B_1} = \frac {B_2C_2}{B_1C_1} \cdot \frac{B_1C_1}{B_2C_2} = 1.$ Combining $\measuredangle C_2A_1D_2$ $=$ $\measuredangle C_2B_2D_2$ $=$ $\measuredangle C_1B_1D_1$ $=$ $\measuredangle C_1A_2D_1$ we get $A_1D_2$ $\stackrel{\parallel}{=}$ $A_2D_1$ $\Longrightarrow$ $A_1X_1$ $\parallel$ $A_2X_2.$

Lemma 2 (well-known) : Given two (not homothetic) triangles $\triangle A_1B_1C_1,$ $\triangle A_2B_2C_2$ . If $L$ is a point s.t. the parallel from $A_2,$ $B_2,$ $C_2$ to $A_1L,$ $B_1L,$ $C_1L$ , resp. are concurrent, then $L$ lies on a circumconic of $\triangle A_1B_1C_1$ or the line at infinity.
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Property 7 : $J,$ $K,$ $J_a,$ $K_a,$ $J_b,$ $K_b,$ $J_c,$ $K_c,$ $P,$ $Q$ lie on a conic.

Proof : From Lemma 1 $\Longrightarrow$ $JJ_a$ $\parallel$ $KK_a,$ $JJ_b$ $\parallel$ $KK_b,$ $JJ_c$ $\parallel$ $KK_c,$ so from Lemma 2 we get the conclusion. $\blacksquare$
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This post has been edited 4 times. Last edited by henderson, Jun 4, 2016, 8:59 PM

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Properties related to the orthocenters of BPC, CPA, APB

by henderson, Jun 4, 2016, 8:54 PM

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