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

The nine-point circle is an amazing circle that passes though the 3 feet of the altitudes and 3 feet of the medians. It also passes though P_A, P_B, and P_C, which are the midpoints of AO, BO, and CO.

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_AM_B is a segment that connects two medians, M_AM_B is parallel to AB. By using similar triangles CQM_B and CH_CA, 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_AM_B. Therefore,

(ii) ΔCH_CM_B is an isosceles triangle

Since triangle ΔCH_CM_B is an isosceles triangle, CM_B = M_BH_C. We know that ∠CAB = ∠CM_BQ because of parallel lines. This also means that ∠CM_BQ = ∠QM_BH_C. Now, the medial triangle is similar to triangle ΔABC. This means that:

(iii) ∠QM_BH_C = ∠M_AM_BH_C = ∠M_CM_AM_B

Since CM_B = M_BA = b/2 and M_AM_C = b/2, the trapezoid H_CM_CM_AM_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_AM_BM_C also passes through H_C.

Similar conclusions can be made about points H_A and H_B.

The Diameter

Triangle ΔH_AH_BH_C (shown in green) is the orthic triangle. A property of orthic triangle holds that ∠H_CH_BP_B = ∠H_AH_BP_B. This becomes useful for proving that M_BP_B is the diameter of the nine-point circle.

We now will prove that M_BP_B is the diameter of the nine-point circle. In the process, we also prove that ΔM_BM_CP_B is a right triangle, and ΔM_BM_AP_B is a right triangle. If these two triangles are right triangles, then M_BP_B must be the diameter of the nine-point circle. To prove that these two are right triangles, we first prove that ΔH_AP_BH_C is an isosceles triangle. We already know that ΔH_AM_BH_C is an isosceles triangle.

Note that ΔH_AH_BH_C is the orthic triangle. Therefore,

(iv) ∠H_AH_BP_B = ∠H_CH_BP_B

This means that H_AP_B = H_CP_B, and triangle ΔH_AP_BH_C is an isosceles triangle. We also have ∠H_AH_BP_B = ∠H_AM_BP_B and ∠H_CH_BP_B = ∠H_CM_BP_B. Since ∠H_AH_BP_B = ∠H_CH_BP_B, ∠H_AM_BP_B = ∠H_CM_BP_B. This proves that M_BP_B is an angle bisector of ∠H_AM_BH_C or that

(v) M_BP_B is a perpendicular bisector of ΔH_AM_BH_C

This means that M_BP_B is the diameter of the nine-point circle and that ΔM_BM_AP_B and ΔM_BM_CP_B are right triangles.

Also, since M_BM_C is perpendicular to AH_A, M_CP_B is parallel to AH_A.

Last Three Points

Our next step is to prove that OP_B = P_BB.

We already know that H_CP_B = H_AP_B. Now, ∠P_AH_AM_C = ∠H_AM_CP_B. This means that:

(vi) P_AM_C = H_AP_B

Since H_BM_C = H_AM_C, ∠H_BP_BM_C = ∠H_AP_AM_C. This proves that the quadrilateral P_AM_CP_BO is a parallelogram. Thus,

(vii) P_AM_C = OP_B

Notice that H_AM_C is a median of ΔAH_AB. As stated earlier, M_CP_B (or M_CM₁') is parallel to AH_A. This proves that

(viii) OP_B = P_BB and M_AM₁' = M₁'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_AP_BP_C is similar to ΔABC. The lengths of the sides of ΔP_AP_BP_C are exactly half that of ΔABC. Therefore, the area of ΔP_AP_BP_C is ¼ 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_AP_BP_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_AP_BP_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.