The Nine-point Circle

28 February 2024

Let us start off with a simple triangle $△ A B C$ . We draw the altitudes for each respective vertex, intersecting the corresponding sides at $H_{A}, H_{B}, H_{C}$ . It is a well-known fact that these three altitudes will intersect at a single point, the orthocenter, denoted as $H$ .

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Next, we take the midpoints of $A H_{A}, B H_{B}, C H_{C}$ and denote them as $T_{A}, T_{B}, T_{C}$ , respectively.

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Finally, the midpoints of $B C, C A, A B$ denoted as $M_{A}, M_{B}, M_{C}$ .

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In case you haven’t known, the points $H_{A}, H_{B}, H_{C}, M_{A}, M_{B}, M_{C}, T_{A}, T_{B}, T_{C}$ all lie inside a common circle, thus the name nine-point circle. Therefore, let’s explore the proof that these nine points indeed lie on a single circle!

The Proof

Let’s focus ourselves on $△ A B H$ . Note that both $M_{C}$ and $T_{B}$ are midpoints of $B A$ and $B H$ respectively. Therefore, $M_{C} T_{B} ∥ A H$ .

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We can use the same idea to prove that $M_{B} T_{C} ∥ A H$ .

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The same idea again to prove that $M_{B} M_{C} ∥ B C$ and $T_{B} T_{C} ∥ B C$ , which also implies $M_{B} M_{C} ∥ T_{B} T_{C} ⟂ A H$ due to the fact that $A H$ is the altitude corresponding to $B C$ . Thus, both $M_{B} M_{C}$ and $T_{B} T_{C}$ are perpendicular to both $M_{C} T_{B}$ and $M_{B} T_{C}$ .

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Combining these arguments, we can conclude that $M_{B} M_{C} T_{B} T_{C}$ is a rectangle!

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Consider the circle that inscribes $M_{B} M_{C} T_{B} T_{C}$ . SInce $M_{B} M_{C} T_{B} T_{C}$ is a rectangle, both $M_{B} T_{B}$ and $M_{C} T_{C}$ are diameters of this circle.

Now, we can repeat these arguments again to prove that $M_{A} M_{B} T_{A} T_{B}$ and $M_{A} M_{C} T_{A} T_{C}$ are also rectangles and are inscribed within this same circle because they have common diameters.

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Therefore, at this point, we managed to prove that $M_{A}, M_{B}, M_{C}, T_{A}, T_{B}, T_{C}$ all lie on the same circle and we’re left with proving that $H_{A}, H_{B}, H_{C}$ also lie on this circle.

This can be done rather quickly because we know that $M_{A} T_{A}$ is a diameter of the circle so for any point $X \neq M_{A}, T_{A}$ in the circle, $∠ M_{A} X T_{A}$ is a right angle. But this is the case for $X = H_{A}$ because $A H_{A} ⟂ B C ∥ H_{A} M_{A}$ . Therefore, $H_{A}$ lies on this circle.

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Analogously we can prove that $H_{B}$ and $H_{C}$ lie on the circle and therefore our proof is complete!

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What’s Next

There some interesting properties derived from the existence of the nine-point circle. You can easily find these online, but my personal favourite would be the Euler line!

The Euler line of $△ A B C$ is a line, that passes through the orthocenter $H$ , the centroid $G$ , the circumcenter $O$ , and the nine-point center $N$ with the following property:

$O G : G N : N H = 2 : 1 : 3$

While the proof might be out of scope, I would like to keep this article short and simple as the main purpose is to appreciate how we can construct such well-worded arguments to proof the existence of the nine-point circle!

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