Orthocenter, Excenter, et al - The Orthocenter, Excenters,...

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Similarly, XC 2245 CY and CF C is the perpendicular bisector of XY , the third side of XYZ . Similarly, ZB 2245 BX and BF B is the perpendicular bisector of ZX , a second side of XYZ . Since AF A BC and BC A , AF A A intersecting at A. Similarly, BF B B intersecting at B and CF C C intersecting at C . A C The Orthocenter, Excenters, Excircles, and the Euler Line (from Sections 4.6 and 4.7) Therefore, the altitudes of ABC , AF A , BF B , and CF C are concurrent at point H . Q E D Therefore, by the "Perpendicular Bisector Concurrence" Theorem, Theorem 4.6.1, the perpendicular bisectors of the sides of XYZ , AF A , BF B , and CF C are concurrent, say at point H. Now, Quadrilaterals ABCY and AZBG are parallograms. By the "Opposite Sides of a Parallogram" Theorem, YA 2245 CB and CB 2245 AZ ; so, YA 2245 AZ by transitivity of congruence. Therefore, AF A is the perpendicular bisector of YZ , one side of XYZ . We show that AF A , BF B , and CF C are the perpendicular bisectors of the sides of XYZ , which are concurrent by the "Perpendicular Bisector Concurrence" Theorem, Theorem 4.6.1. B F C F A F B Theorem 4.6.4, The "Altitude Concurrence" Theorem: The three altitudes of a triangle ABC are concurrent at a point H . The point of concurrence H is called the Orthocenter of the triangle. Proof: Let ABC be any triangle. Construct lines A , B , and C , through A, B, and C and parallel to the side opposite these points, respectively. Let
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Orthocenter, Excenter, et al - The Orthocenter, Excenters,...

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