Ortho-circumcentric complexity

Consider an acute-angled triangle $ABC$ , with circumcenter $O$ .

Denote the point $D$ such that it lies on segment $BC$ .
Let $O_{1}$ and $O_{2}$ be the circumcentres of triangle $ABD$ and $ACD$ , respectively.

Consider the triangle $O_{1}O_{2}D$ , with orthocenter $H_1$ .

Prove that: $O H_{1}$ is parallel to $BC$ .

Bonus: Prove this result for an obtuse-angled triangle as well.

#Geometry

Easy Math Editor

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Comments

For a general triangle $\overline{ABC}$ , we define $p_{ij}$ to be the perpendicular bisector of the two points $i$ and $j$ , and $h_{ij}$ to be the altitude of the triangle corresponding to the line through two points $i$ and $j$ of said triangle. So, the perpendicular bisectors $p_{AB}$ , $p_{AC}$ and $p_{BC}$ are perpendicular to $AB$ , $AC$ and $BC$ respectively and all meet at the circumcentre $O$ of $\overline{ABC}$ . For a point $D$ on $BC$ , we may suppose that $O_1$ and $O_2$ are the circumcentres of $\overline{ABD}$ and $\overline{ACD}$ . We then have that $O_1$ and $O_2$ are respectively the meets of the concurrent triple of lines $(p_{AB},p_{AD},p_{BD})$ and $(p_{AC},p_{AD},p_{CD})$ . As $p_{AD}$ passes through both points, we can deduce that $p_{AD} = O_1O_2$ . For the triangle $\overline{O_1O_2D}$ , we then have that the altitude $h_{O_1O_2}$ of this triangle is simply $p_{AD}$ ; this then implies that the orthocentre $H_1$ of $\overline{O_1O_2D}$ lies on $AD$ . It remains to show that $H_1O$ is parallel to $BC$ .

I will get back to you on this, as I will need a couple more days to work this out.

A Former Brilliant Member - 2 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$