# Nine-Point Circle

One of the greatest occurrences in geometry is the Nine-Point Circle. This exercise will walk you through the proof of the Nine-Point Circle.

## The Nine-Point Circle Explained

## Proof of the Altitudes

First, we'll prove that the cirlce that passes through the feet of the medians also passes through the feet of the altitudes. In other words, if we draw a circle that passes through the feet of the medians, then it will also pass through the feet of the altitudes. We'll use H_{C} as an example and the others can be proved similarly.

We connect the points M_{B} and H_{C}. We first prove that CQ = QH_{C}. Since M_{A}M_{B} is a segment that connects two medians, M_{A}M_{B} is parallel to AB. By using similar triangles CQM_{B} and CH_{C}A, it is easy to prove that:

(i) CQ = QH_{C}

Also, since CH_{C} is an altitude of triangle ABC, CQ is perpendicular to segment M_{A}M_{B}. Therefore,

(ii) ΔCH_{C}M_{B} is an isosceles triangle

Since triangle ΔCH_{C}M_{B} is an isosceles triangle, CM_{B} = M_{B}H_{C}. We know that ∠CAB = ∠CM_{B}Q because of parallel lines. This also means that ∠CM_{B}Q = ∠QM_{B}H_{C}. Now, the medial triangle is similar to triangle ΔABC. This means that:

(iii) ∠QM_{B}H_{C} = ∠M_{A}M_{B}H_{C} = ∠M_{C}M_{A}M_{B}

Since CM_{B} = M_{B}A = b/2 and M_{A}M_{C} = b/2, the trapezoid H_{C}M_{C}M_{A}M_{B} is an isosceles trapezoid. We know that the vertices of an isosceles triangle lie on a circle. Thus, the circle that passes through ΔM_{A}M_{B}M_{C} also passes through H_{C}.

Similar conclusions can be made about points H_{A} and H_{B}.

## The Diameter

Triangle ΔH

_{A}H

_{B}H

_{C}(shown in magenta) is the orthic triangle. A property of orthic triangle holds that ∠H

_{C}H

_{B}P

_{B}= ∠H

_{A}H

_{B}P

_{B}. This becomes useful for proving that M

_{B}P

_{B}is the diameter of the nine-point circle.

We now will prove that M_{B}P_{B} is the diameter of the nine-point circle. In the process, we also prove that ΔM_{B}M_{C}P_{B} is a right triangle, and ΔM_{B}M_{A}P_{B} is a right triangle. If these two triangles are right triangles, then M_{B}P_{B} must be the diameter of the nine-point circle. To prove that these two are right triangles, we first prove that ΔH_{A}P_{B}H_{C} is an isosceles triangle. We already know that ΔH_{A}M_{B}H_{C} is an isosceles triangle.

Note that ΔH_{A}H_{B}H_{C} is the orthic triangle. Therefore,

(iv) ∠H_{A}H_{B}P_{B} = ∠H_{C}H_{B}P_{B}

This means that H_{A}P_{B} = H_{C}P_{B}, and triangle ΔH_{A}P_{B}H_{C} is an isosceles triangle. We also have ∠H_{A}H_{B}P_{B} = ∠H_{A}M_{B}P_{B} and ∠H_{C}H_{B}P_{B} = ∠H_{C}M_{B}P_{B}. Since ∠H_{A}H_{B}P_{B} = ∠H_{C}H_{B}P_{B}, ∠H_{A}M_{B}P_{B} = ∠H_{C}M_{B}P_{B}. This proves that M_{B}P_{B} is an angle bisector of ∠H_{A}M_{B}H_{C} or that

(v) M_{B}P_{B} is a perpendicular bisector of ΔH_{A}M_{B}H_{C}

This means that M_{B}P_{B} is the diameter of the nine-point circle and that ΔM_{B}M_{A}P_{B} and ΔM_{B}M_{C}P_{B} are right triangles.

Also, since M_{B}M_{C} is perpendicular to AH_{A}, M_{C}P_{B} is parallel to AH_{A}.

## Last Three Points

Our next step is to prove that OP_{B} = P_{B}B.

_{C}P

_{B}= H

_{A}P

_{B}. Now, ∠P

_{A}H

_{A}M

_{C}= ∠H

_{A}M

_{C}P

_{B}. This means that

(vi) P_{A}M_{C} = H_{A}P_{B}

Since H_{B}M_{C} = H_{A}M_{C}, ∠H_{B}P_{B}M_{C} = ∠H_{A}P_{A}M_{C}. This proves that the quadrilateral P_{A}M_{C}P_{B}O is a parallelogram. Thus,

(vii) P_{A}M_{C} = OP_{B}

Notice that H_{A}M_{C} is a median of ΔAH_{A}B. As stated earlier, M_{C}P_{B} (or M_{C}M_{1}') is parallel to AH_{A}. This proves that

(viii) OP_{B} = P_{B}B and M_{A}M_{1}' = M_{1}'B.

## The Radius of the Nine-point Circle

Now we wonder about the length of the radius of the nine-point circle. It turns out that the radius is exactly half of the radius of the circumcircle of ΔABC. First, we note that ΔP_{A}P_{B}P_{C} is similar to ΔABC. The lengths of the sides of ΔP_{A}P_{B}P_{C} are exactly half that of ΔABC. Therefore, the area of ΔP_{A}P_{B}P_{C} is ΒΌ of ΔABC. The circumradius, *R*, of ΔABC is given by: $R=\frac{abc}{4K}$, where *K* is the area of ΔABC. The circumradius of ΔP_{A}P_{B}P_{C} (denoted as *R*_{N}) is given by: $R_{N}=\frac{\frac{a}{2}\cdot\frac{b}{2}\cdot\frac{c}{2}}{4\cdot\frac{1}{4}\cdot K}=\frac{abc}{8K}=\frac{1}{2}R$.

## Medial Triangle Similarity

Triangle ΔP_{A}P

_{B}P

_{C}is also similar to the medial triangle. Its sides are parallel to ΔABC.

Moreover, when its vertices are connected with the medians, the lines are parallel to the altitudes (this is shown in the figure above). These lines meet the sides halfway between the feet of the altitudes and the vertices.