The circle $\omega_1$ intersects the circle $\omega_2$ in the points $A$ and $B$, a tangent line to this circles intersects $\omega_1$ and $\omega_2$ in the points $E$ and $F$ respectively. Suppose that $A$ is inside of the triangle $BEF$, let $H$ be the orthocenter of $BEF$ and $M$ is the midpoint of $BH$. Prove that the centers of the circles $\omega_1$ and $\omega_2$ and the point $M$ are collinears.
Problem
Source: Lusophon MO
Tags: geometry
pablock
18.02.2018 17:43
Let $\{N\}=\overleftrightarrow{BA} \cap FE$, $\{C\}=\overleftrightarrow{FA} \cap BE$, $\{D\}=\overleftrightarrow{AE} \cap FB$, $\{J\}=\overleftrightarrow{FH} \cap BE$, $\{K\}= \overleftrightarrow{HE} \cap FB$, $O_1$, $O_2$ be the centers of $w_1$, $w_2$ and $\{P\}= BA \cap O_2O_1$.
Note that $NE^2=AN \cdot BN=FN^2 \implies NE=FN$, so $N$ is the midpoint of segment $FE$.
By Ceva's theorem, since $DE$, $BN$ and $FC$ concur,
$\frac{FN}{NE} \cdot \frac{CE}{BC} \cdot \frac{DB}{FD}=1 \implies \frac{BC}{CE}=\frac{DB}{FD} \iff \overleftrightarrow{DC} \parallel \overleftrightarrow{FE}$.
Thus $\angle ADC = \angle AEF$. Since line $FE$ is tangent to $w_1$ at $E$, we have $\angle AEF = \angle ABC = \angle ADC$, so $ADBC$ is cyclic. Then $\angle EDB = 90^{\circ}+ \angle AEH = \angle FCE = 90^{\circ}+AFH$ $\implies \angle AEH=\angle AFH \implies HAEF$ is cyclic $\implies \angle AEF = \angle AHJ = \angle ABE = \angle ABJ \implies BCAH$ is cyclic $\implies \angle BAH = \angle BJH = 90^{\circ}$.
Moreover, $BA \bot O_2O_1$ and $P \in O_2O_1$. Because both $M$ and $P$ are midpoints, $MP \parallel HA \implies \angle BPM =\angle BAH = 90^{\circ} = \angle BPO_2 \implies M \in \overleftrightarrow{O_2P} = \overleftrightarrow{O_2O_1}$, then we are done.
ShinyDitto
18.02.2018 20:17
$\angle EAF= 180- \angle EBF = \angle EHF \longrightarrow EAHF$ cyclic. $\angle BAH = 360 - \angle BAE - \angle EAH = 360 - (180 - \angle BEF) - (180- \angle HFE) =90$ $\rightarrow M$ lies on the perpendicular bisector of $BA$, which happens to be the line joining the centers of $w_1$ and $w_2$.
trying_to_solve_br
11.08.2021 17:24
Mehhhhhhh Notice that $A$ is the $B-$humpty point of $BEF$ (by definition), and it is well known that $HA\perp AB$, since by PoP $AB$ passes through the midpoint of $EF$. Then, since the midpoint of $BH$ is the circumcenter of $ABH$, and by coaxal circles the three centers are collinear we finish the problem.