Help on this Geometry Problem!

Prove that in $\triangle ABC$ , circle $ABC$ is the nine-point circle of the extracentral triangle of $ABC$ , that is, the triangle whose vertices comprise of the excenters.

You are free to share your proof below in the comments section.

#Geometry

Easy Math Editor

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`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`	$\sum_{i=1}^3$
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Comments

$\triangle ABC$ is orthic triangle of the triangle whose vertices are ex-centers. To reverse the scenario, if we look at the orthic triangle of any given acute triangle, its angle bisectors are altitudes and sides of the reference triangle. This means that circumcirle of the orthic triangle is nine point circle of the reference triangle.

Maria Kozlowska - 5 years, 8 months ago

Wait how do you know that $\triangle ABC$ is the orthic triangle?

Alan Yan - 5 years, 8 months ago

Carefully read the second statement and make a good drawing. You can also check this website: orthic triangle

Maria Kozlowska - 5 years, 8 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}`	$\boxed{123}$