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14 CHAPTER 1 ADVANCED EUCLIDEAN GEOMETRY Orthocel In the trian- B. X gle AAB nes (AX), (BY), and Z (CZ) are wn so that (AX) _ (BC), (BY - (CA), and (CZ) 1 (AB). Cles we either have C AZ Y CY y ZB' YA B or that exactl le of these ratios is positive. We re C ly AABY ~ AAC AZ CZ AY BY A X Likewise, we hav BX AX AABX ~ AC BZ CZ and ACBY ~ ACAX CY BY CX AX Therefore, AZ BX CY BX CY CZ AX BY ZB XC YA BZ CX BY CZ AX . = 1. By Ceva's theorem the line 1X), (BY), and (CZ) are concurrent, and the point of concurrency illed the orthocenter of AABC. (The line segments [AX], [BY]. 1 [CZ] are the altitudes of AABC.) Incenter. In the triangle IBC lines B that (AX ) bisects DAY, ABC, and (CZ) bisects A As we show below, that the lines (), (BY), and (CZ) are concurrent; 3 point of concurrency is called the center of AABC. (A very interesti "extremal"CHAPTER 1 ADVANCED EUCLIDEAN GEOMETRY AB - AA. Let B P' be the point determine bisector (BP') of ABC. Then by Win has already been proved above, we have Two. But this implies that AP AP' PC PC P =P. A Conclusion of the proof that gle bisectors are concurrent. First of all, it is clear that the r ant ratios are all positive. By the Angle Bisector Theorem, AB AY BZ AB BX BC YO' ZA' AC XC' therefore, CY CA AB BC xc X YA BC X AC X = 1. AB Ceva's Jrem now finishes the job! EXER ES 1. TL ngle Bisector Theorem involved the bisection of one of the give iangle's interior angles. Now let P be a point on the line (AC ternal to the segment [AC]. Show that the line (BP) bisect: external angle at B if and only if AB AP BC - PC 2. You are given the triang- " ha the point of inter- section of the bisector of BAC with jew ) be the point of intersection of the bisector of CBA with [AC]. ally, let Z be the point of intersection of the exterior angle bise the line (AB). Show that X, Y, and Z are colinear."