Given an acute triangle $ABC$ ($AC\ne AB$) and let $(C)$ be its circumcircle. The excircle $(C_1)$ corresponding to the vertex $A$, of center $I_a$, tangents to the side $BC$ at the point $D$ and to the extensions of the sides $AB,AC$ at the points $E,Z$ respectively. Let $I$ and $L$ are the intersection points of the circles $(C)$ and $(C_1)$, $H$ the orthocenter of the triangle $EDZ$ and $N$ the midpoint of segment $EZ$. The parallel line through the point $l_a$ to the line $HL$ meets the line $HI$ at the point $G$. Prove that the perpendicular line $(e)$ through the point $N$ to the line $BC$ and the parallel line $(\delta)$ through the point $G$ to the line $IL$ meet each other on the line $HI_a$.
Problem
Source: BMO Shortlist 2017 G8
Tags: geometry, excircle, concurrent, concurrency, circumcircle
Pluto04
14.07.2019 18:49
Bumppppp
Pluto04
15.07.2019 15:47
Bumppppp
andrenguyen
15.07.2019 17:53
Let $J, N, K$ are the midpoints of 3 segments $DE, EZ, DZ$ respectively. Let $M$ is the circumcentre of triangle $HEZ$. $F$ is the centre of the nine-point circle of triangle $DEZ$. Hence $F$ is the midpoint of $HI_a$ and also the midpoint of $MD$. Moreover, we have $N$ is the midpoint of $MI_a$ hence $NF \parallel I_aD$ hence $NF \perp BC$. hence $NF \equiv (e)$ Label that $(ABC)$ is the circumcircle of triangle $ABC$. Now let $\Gamma$ is the inversion with centre $I_a$ and the power $ID^2$. We have $\Gamma: N \leftrightarrow A, J \leftrightarrow B, K \leftrightarrow C, (ABC) \leftrightarrow (MNJ)$ We also have $(ABC) \cap (MNJ) = \lbrace I, L \rbrace$. Therefore $\Gamma: I \leftrightarrow I, L \leftrightarrow L$ Hence $F, O, I_a, H$ lie in the perpendicular bisector line of $IL$. Therefore $\angle GI_aH = \angle I_aHL = \angle I_aHG$ so $GF \perp HI_a$ hence $GF \parallel IL$ hence $GF \equiv (\delta)$ Therefore, $(\delta), (e), I_aH$ are concurrent at point $F$.
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