Question

Let H be the orthocenter of a nondegenerate...

14. Let H be the orthocenter of a nondegenerate & ABC. Prove that the second point of intersection of two circles with diameters (AH] and [BH] lies on the line (AB).

Answer 1

poin F say. SABC on non-clapenerale ie A; B and C ar difline and non-collinear periods. Let D, E and F be foot be attchides from vertex A, B and C onto segments, BC, AC and AB respectively. Also let AD BE on CF are Concurrent at poor tie. His orthocenter B Let the circle with diameter [Ah] inferseets AB ar AFH sublended by diameter E H ДАВС of D Then, = 90° because oogle at any point on me remaining cich is 90° Therefore HF! | AF & HF I AB (: A-F'- B 1.c. A, FB are collincar in that order. Also, 7 ĈE 1 AB ( CF in alihide from conto AB) HF L AB 1: CF passes through Orthovenier H ie C-H-F) Siru, perpurdicelor from a pono to a line en is unique, therefore HF coincide with HF! (from 0 ond®), F=F

Similarly, Henu, circle with diameter [Ah] pase through the point F (foot of allinud from C onito AB AB). circle with diamero [BH] passes through point F (foot of alhlude from Conra AB). point of is the second point of intersection of the Hence h Two circles with chiameters [Ah] and [BH]. Sino port f is he foor of me altrude from C onto AB, therefore I lies on AB. So me posor of o intersection of the two circles with diameter [Ah] and [BH] lie on line (AB). (The point pont of intersection me foot of altitude from C.) GED Second being