![]() 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. ![]() ![]()
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