Nine-point circle - proof using plane geometry

Question

I am taking a course in multivariable calculus this year & I thought it would be a good idea to brush up plane and solid geometry.

I would like to prove that, for any given triangle, there is a unique circle that passes throught the three midpoints of sides, the feet of the altitudes, and the middle points on the segment joining the feet of the perpendicular and the orthocenter. Could you please give me any hints, that would lead to the correct proof?

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For an example of the type of work I've done ... On some results related to triangle centers, I was able to prove that the centroid $G$ divides the line joining the orthocenter $H$ and the circumcenter $K$ in the ratio $2:1$. I wasn't sure how to prove that $H$, $G$, $K$ are collinear. I learnt from Kiselev's Geometry book about how to prove this result. The idea is that the circumcenter of the original triangle is the orthocenter of the medial triangle $H^\prime$. And the medial triangle reflected about the centroid $G$, followed by dilation of a factor $2:1$ results in the original triangle. Thus, the center $H^\prime$ moves to the point $H$. Both triangles share the centroid $G$. So, $H$, $G$, $K$ are collinear.

Cheers, Quasar.

Answer 1

The nine point circle exists in all triangles , but for this answer , we will be focusing on the case when $\triangle ABC$ is acute .

First , let us remember the properties of cyclic quadrilaterals , as this will help us prove the concyclicity of points .

The main points to remember , are that the opposite angles are supplementary , and that the chords subtend equal angles at the circumference . The equal point to note is that the converse is true as well. This is how we will be proving that the points lie on a circle.

For the problem , it may be best to solve it in steps .

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Step $1$:-

In step $1$ , we can perhaps prove that the feet of the altitudes lie on the same circle as the midpoints.

To do this , a single altitude should suffice . As, if we can prove that this lies on the same circle as the midpoints , the others must as well. To prove this , maybe the midpoint theorem , and basic angle chasing will help...

Hint:-

Prove $\triangle FAX$ is isosceles using the converse of the midpoint theorem . Then prove that $\angle FXE$ and $\angle FDE$ are supplementary .

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Step $2$:-

I assume you have finished step $1$ . Now , we use the information that the feet of the altitudes and the midpoints lie on the same circle , to prove that the midpoints of the lines joining the vertices to the orthocentre also lie on the same circle . Again , properties of altitudes and midpoints should help us solve this . Note that in the figure , $M’$ is the midpoint of the line joining vertex $C$ to orthocentre $H$ . $M$ is the midpoint of $BC$

Hint:-

$M’M$ joins the midpoints of two line segments ! Use the midpoint theorem . Also , make a great observation. Since $\angle CFB = \angle BEC = 90 $ , $CB$ is the diameter of the circle which inscribes cyclic quadrilateral $CEFB$ ! Also , $M$ is the midpoint of the diameter... By angle chasing , and using these observations , you should be able to prove that $\angle FM’M = \angle FEM$, proving that the quadrilateral is cyclic

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This completes our proof , as it must follow that all the other midpoints of the lines joining the vertices with the orthocentre are concyclic as well.

Ofcourse , we have proved the existence of the $9$-point circle for only acute angled triangles , but I trust you should now be able to prove it for all other triangles as well!

Answer 2

An alternate method of proof utilises the property of dilations (or homothety).

Firstly, consider reflecting $H$ over $D$. You get a new point $H'$. Show that $H'$ lies on circle $ABC$.

$\angle BH'C = \angle BHC = 180^{\circ} - (\angle HBC + \angle HCB) = 180^{\circ} - (\angle HAC + \angle HAB)$ since $AEDB$ and $AFDC$ are cyclic. However, the sum of these two angles is simply $\angle BAC$, so $\angle BAC + \angle BH'C = 180^{\circ}$, so $ABH'C$ is cyclic.

Using this for $E$ and $F$ gives you similar results. Now consider what happens when you reflect $H$ over $L$. Call this point $H''$. Show that $H''$ lies on circle $ABC$.

Since $L$ is the midpoint of $BC$ and $HH''$, we must have $HBH''C$ is a parallelogram! Now, we get $\angle BH''C = \angle BHC = \angle BH'C$ from before, so we must have $H''$ also on circle $ABC$.

Using this for $M$ and $N$ gives similar results yet again.

Now, finally consider what happens if you dilate the entire circle $ABC$ through point $H$ with factor $\frac{1}{2}$.

All the points on the circle go to the midpoints of $H$ and the original point of the circle. Thus, $H'$ goes to $D$, $H''$ goes to $L$, and $A$ goes to $I$.

This holds for all other points, so they must all lie on a circle - namely the nine-point circle.

Hopefully this helps, but you can ask further queries if a part of my proof is a bit too obscure.