Similarly,XC2245CYandCFCis the perpendicular bisector ofXY,the third side ofXYZ .Similarly,ZB2245BXandBFBis the perpendicular bisector ofZX,a second side ofXYZ .SinceAFA⊥BCandBCA,AFA⊥Aintersecting atA.Similarly,BFB⊥Bintersecting atBandCFC⊥Cintersecting atC .ACThe Orthocenter, Excenters, Excircles, and the Euler Line(from Sections 4.6 and 4.7)Therefore, the altitudes ofABC ,AFA,BFB,andCFCare concurrent at point H .Q E DTherefore, by the "Perpendicular Bisector Concurrence" Theorem, Theorem 4.6.1,the perpendicular bisectors of the sides ofXYZ ,AFA, BFB, and CFCare concurrent,say at point H.Now,Quadrilaterals ABCY and AZBG are parallograms.By the "Opposite Sides of a Parallogram" Theorem,YA2245CBandCB2245AZ;so,YA2245AZby transitivity of congruence.Therefore,AFAis the perpendicular bisector ofYZ,one side ofXYZ .We show thatAFA, BFB, and CFCare theperpendicular bisectors of the sides ofXYZ ,which are concurrent bythe "Perpendicular BisectorConcurrence" Theorem, Theorem 4.6.1.BFCFAFBTheorem 4.6.4, The "Altitude Concurrence" Theorem:The three altitudes of a triangle ABC are concurrentat a point H .The point of concurrenceHis called the Orthocenter of the triangle.Proof:LetABC be any triangle.Construct linesA,B, andC,through A, B, and C and parallel to the side opposite these points, respectively.LetX , Y and Zbe the points whereBandC,AandC, andAandCintersect in pairs, respectively.