Let CO'D cut the circle (O') at E. By the result of Problem 719, E is the incentre of ∆ABC Observing that ED is a diameter of circle (O'), it follows that AD ⊥ AE, implying that AD is the external bisector of ∠CAB. By symmetry, BD is the exteral bisector of ∠CBA. Hence D is an excentre of ∆ABC
Let CO'D cut the circle (O') at E.
ReplyDeleteBy the result of Problem 719,
E is the incentre of ∆ABC
Observing that ED is a diameter of circle (O'),
it follows that AD ⊥ AE, implying that
AD is the external bisector of ∠CAB.
By symmetry,
BD is the exteral bisector of ∠CBA.
Hence D is an excentre of ∆ABC