12/22/2023 0 Comments Altitude geometry proofs![]() Continuingwith the same figure, angle ABB* = angle AA*B*. Thus the circle intersects AC at a point P so that BP is perpendicularto AC the only such point is P = B*. By the inscribed angle theorem (Carpenter theorem),since AC'B is a diameter and a straight angle, for any point P on c 3,the angle APB is a right angle. The center of the circleis the midpoint C' of AB. Continuing with the same figure, thecircle c 3 with diameter AB intersects AC at B* and BC as A*. Part 1: Prove that the altitudes and sides of ABC are angle bisectors ofA*B*C* ![]() From this it can be proved that the orthic triangle A*B*C* has the smallest perimeter of any triangle with vertices on the sides of ABC.The sides of the orthic triangle form an "optical" or "billiard" path reflecting off the sides of ABC.The altitudes and sides of ABC are interior and exterior angle bisectors of orthic triangle A*B*C*, so H is the incenter of A*B*C* and A, B, C are the 3 ecenters (centers of escribed circles).Thistriangle has some remarkable properties that we shall prove: The orthic triangle of ABC is defined to be A*B*C*. ![]() (Theproof used the relationship between the perpendicular bisectors of the sides ofa triangle and the altitudes of its midpoint triangle). Then we have proved earlier that the altitudes areconcurrent at a point H. Given a triangle ABC with acute angles, let A*, B*, C* bethe feet of the altitudes of the triangle: A*, B*, C* are points on the sidesof the triangle so that AA* BB*, CC* are altitudes. Altitudes and the Orthic Triangle of Triangle ABC
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