The altitudes $AH, CH$ of an acute-angled triangle $ABC$ meet the internal bisector of angle $B$ at points $L_1, P_1$, and the external bisector of this angle at points $L_2, P_2$. Prove that the orthocenters of triangles $HL_1P_1, HL_2P_2$ and the vertex $B$ are collinear.
Problem
Source: Sharygin Finals 2018 Grade 10 P1
Tags: geometry
01.08.2018 09:57
Let $AH \cap BC \equiv D$ , $CH \cap AB \equiv F$.Let the internal,external bisector of angle $B$ meet $\odot BDF$ at $M,N$ resp.Let $H_1,H_2$ be the orthocenters of $\triangle HL_1P_1,\triangle HL_2P_2$ resp. We make the following observations step by step(proving it is a matter of simple angle/ratio chasing), 1)$BMHN$ is a rectangle with circumcircle $\odot BDF$ 2)$M-H_1-H$ and $N-H-H_1$ are collinear in this order 3)$\triangle MP_1H_1 \sim \triangle NP_2H$ and $\triangle MP_1H \sim \triangle NP_2H_2$ 4)$\frac{MH_1}{H_1H} = \frac{NH}{HH_2}$ 5)$B-H_1-H_2$ is collinear
29.07.2022 13:43
The second official solution is quite clever. Just use the fact that the $4$ orthocenters formed by any four lines in the plane (in general position) are collinear. Take the four lines to be the two altitudes and two angle bisectors. $B$ is the orthocenter of two of the triangles (as two bisectors are perpendicular). So we are done. Remark: The above proof actually shows that the two angle bisectors can be replaced by any two perpendicular lines passing through $B$.