"Let $Q$ $\equiv$ $(BX_CX_A)$ $\cap$ $(CX_AX_B)$. We have $(QB, QC) \equiv (QB, QX_A) + (QX_A, QC) \equiv (X_CB, X_CX_A) + (X_BX_A, X_BC) \equiv (AB, AD) + (AD, AC)$ $\equiv (AB, AC) \pmod \pi$. Then $Q$ $\in$ $(ABC)$. Similarly, if we let $Q'$ $\equiv$ $(CX_AX_B)$ $\cap$ $(AX_BX_C)$ then $Q'$ $\in$ $(ABC)$. Hence $Q'$ $\equiv$ $Q$.Let $S$ $\equiv$ $(CAX_B)$ $\cap$ $(ABX_C)$. We have $(SX_B, SX_C) \equiv (SX_B, SA) + (SA, SX_C) \equiv (CX_B, CA) + (BA, BX_C) \equiv (CX_B, CQ) + (CQ, CA)$ $+ (BA, BQ) + (BQ, BX_B) \equiv (X_AX_B, X_AQ) + (X_AQ, X_AX_C) \equiv (X_AX_B, X_AX_C) \pmod \pi$. Then $S$ $\in$ $(X_AX_BX_C)$. Similarly, if let $S'$ $\equiv$ $(ABX_C)$ $\cap$ $(BCX_A)$ then $S'$ $\in$ $(X_AX_BX_C)$. Hence $S'$ $\equiv$ $S$.", khanhnx.
Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...
"Let $Q$ $\equiv$ $(BX_CX_A)$ $\cap$ $(CX_AX_B)$. We have $(QB, QC) \equiv (QB, QX_A) + (QX_A, QC) \equiv (X_CB, X_CX_A) + (X_BX_A, X_BC) \equiv (AB, AD) + (AD, AC)$ $\equiv (AB, AC) \pmod \pi$. Then $Q$ $\in$ $(ABC)$. Similarly, if we let $Q'$ $\equiv$ $(CX_AX_B)$ $\cap$ $(AX_BX_C)$ then $Q'$ $\in$ $(ABC)$. Hence $Q'$ $\equiv$ $Q$.
ReplyDeleteLet $S$ $\equiv$ $(CAX_B)$ $\cap$ $(ABX_C)$. We have $(SX_B, SX_C) \equiv (SX_B, SA) + (SA, SX_C) \equiv (CX_B, CA) + (BA, BX_C) \equiv (CX_B, CQ) + (CQ, CA)$ $+ (BA, BQ) + (BQ, BX_B) \equiv (X_AX_B, X_AQ) + (X_AQ, X_AX_C) \equiv (X_AX_B, X_AX_C) \pmod \pi$. Then $S$ $\in$ $(X_AX_BX_C)$. Similarly, if let $S'$ $\equiv$ $(ABX_C)$ $\cap$ $(BCX_A)$ then $S'$ $\in$ $(X_AX_BX_C)$. Hence $S'$ $\equiv$ $S$.", khanhnx.