올림피아드 문제 풀이 중에... 아래에 밑줄친 부분의 뜻이 뭔가요?
[Problem] : Let AE and BF be altitudes, and H the orthocentre, of acute triangle ABC. The reflection of AE across the interior angle bisector of A meets the reflection of BF across the interior angle bisector of B meet in a point O. The lines AE and AO meet the circumcircle of ABC again at M and N, respectively. Let P, R, S be the intersection of BC with HN, BC with OM, HR with OP, respectively. Show that AHSO is parallelogram.
[Solution] : Note that O and H are isogonal conjugates, where H is orthocentre of ABC and O is centre of circumcircle of ABC.
First we shall prove that AO || HR. AO = OM = R. It's easy to see that HE = EM. (<HBE = < EBM = 90 - <c)
So <AOM = < OMA = <RMH = < RHM => AO || HR.
AN = 2R, and <AMN = 90. <AEP = 90, and we conclude that EP || MN. From HE = EM we get that P is midpoint of line NH. O is midpoint of AN so OP || AH in triangle AHN.
AO || HR and OP || AH => AHSO is paralelogram.