In the acute triangle $ABC, \angle ABC = 60^o , O$ is the center of the circumscribed circle and $H$ is the orthocenter. The angle bisector $BL$ intersects the circumscribed circle at the point $W, X$ is the intersection point of segments $WH$ and $AC$ . Prove that points $O, L, X$ and $H$ lie on the same circle.
Problem
Source: 2019 Oral Moscow Geometry Olympiad grades 8-9 p3
Tags: geometry, circumcircle, Concyclic, Circumcenter, orthocenter
Pi-is-3
21.05.2019 22:29
What is L? You haven't said what point is L. Is it any random point?
parmenides51
21.05.2019 22:32
$L$ is the intersection point of the angle bisector of $BL$ with the opposite site $AC$ (it felt obvious to me)
AlastorMoody
23.07.2019 23:30
Let $H'$ be reflection of $H$ over $\overline{AC}$. Since, $\angle ABC=60^{\circ}$ $\implies$ $O$ $\in$ $\odot (AHC)$ $\implies$ $O$ is reflection of $W$ over $\overline{AC}$ $\implies$ $H'XLW$ is reflection of $HXLO$ over $\overline{AC}$ $\implies$ $HW$ $\cap$ $OH'$ $=$ $X$ $\in$ $\overline{AC}$ $\implies$ $\angle XH'W$ $=$ $90^{\circ}-\angle XBL$ $=$ $\angle BLX$ $\implies$ $H'XLW$ is cyclic $\implies$ $HXLO$ is cyclic
Mahdi_Mashayekhi
29.12.2021 18:41
Let H' be reflection of H about AC. ∠AOC = 120 = ∠AWC and AO = CO , WA = WC so W is reflection of O about AC so it's enough to prove XLWH' is cyclic. HOWH' is isosceles trapezoid sand AC is perpendicular bisector of both HH' and OW so if H,X,W are collinear then O,X,H' are collinear as well. ∠BLX = 90 - ∠HBL and ∠WH'O = 90 - ∠H'BW so ∠BLX = ∠WH'O and XLWH' is cyclic. we're Done.