# Assignment 4

## 9 Point Circle

Kate Berryman
cavaleri@uga.edu

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The Nine-Point circle for any triangle passes through the three mid-points of the sides, the three feet of the altitudes, and the three mid-points of the segments from the respective vertices to orthocenter.

To start, we begin with some constructions:

 Take an arbitrary triangle ABC. Take M to be the midpoint of AB, N be the midpoint of BC, and L be the midpoint of AC. Next, take D to be the foot of the height from C to AB, E to be the foot of the height from A to BC, and F to be the foot of the height from B to AC. Here, we see that point O is our orthoceenter. Last, take P to be the midpoint of OB, Q be the midpoint of OC, and R to be the midpoint of OA.

How do we prove that points D, E, F, M, N, L, P, Q, and R lie on a circle?

Proof:

 For the first part of this proof, we must remember the theorem that states if a line bisects one side of a triangle and bisects another side of the same triangle, then this line is parallel to the third side of the triangle Since L bisects AC and N bisects BC, then the line LN is parallel to AB. Similiarly, in triangle AOB, R bisects AO and P bisects BO, therefore RP is paralle to AB. Now, lets look at the right triangle ACD. If we drop the foot of the perpendicular on LD to get point T, we see that T bisects AD. Therefore LT is parallel to CD and LR is parallel to CD. Similiarly, CD is parallel to NP Therefore, NP is parallel to LR. Since LR is parallel to NP and LN is parallel to NP, we have the quadrilateral LNPR, in particular a rectangle with all angles having a measure of 90. By our previously stated theorem, we have that LM is parallel to BC. Now lets look at triangle BCO. We know that Q bisects OC and P bisects OB, therefore QP is parallel to BC.It follows that LM is parallel to QP. Now, look at triangle AEB. We can see that ABE is a right triangle. Let J be the foot of the perpendicular from M to BE. By our previously stated theorem, MJ is parallel to AE. So, MP is parallel to AE. Similiarly, look at right triangle AEC. Let K be the foot of the perpendicular dropped from L to CE. Then, LK is parallel to AE. So, LQ is parallel to AE. Since LM is parallel to QP and LQ is parallel to MP, then LMQP is a parallelogram, in particular LMQP is a rectangle since each angle has a measure of 90. Now, we see two rectangles in our triangle ABC. Since LNPR is rectangle, the diagonals LP and NR bisect and are congruent. So LP = NR and they intersect at G. Therefore, LG = PG = NG = RG. Similiarly, in rectangle LQMP has diagonals LP and MQ that bisect at H. Since LQMP is a rectangle LP = MQ and LH = PH = MH = QH. Since LP= NR and LP = MQ then NR = MQ. They intersect G and H in LNPR, so G = H. Since LQPM and LNPR are rectangles, so it follows that LG = PG = NG = RG and LH = PH = MH = QH. Since G = H, then LG = PG = MG = QG. So, LG = PG = NG = RG = MG = QG. By definition of a circle, these are six radii with G being the center and L, P, N, R, M, and Q being points the circle. Take the right triangle DQM, with D being the foot of the perpendicular. We see the side QM = QG + GM. We also know that GQ = GM with G as the midpoint. We know by the previously stated theorem that if G is a midpoint of the hypotenuse of right triangle DQM, the G is equidistance from the vertices D, Q, M and a circle passing through D, Q, M has G as its circumcenter. Therefore, D is on the perimeter of the circle. In the same way, lets look at the right triangle LFP. We know that LP = LG + GP and LG = GP. By the previously stated theorem, G is the circumcenter of our circle that has points L, F, P on the perimeter. Therefore, this shows that F is on the perimeter. Similiarly, let's look at the right triangle NER, where NR = NG + GR and we know that NG = GR. By our theorem, G again is the circumcenter with points N, E, R on its perimeter giving us our ninth point E. Therefore, G is the center of a nine point circle D, E, F, M, N, L, P, Q, and R. So, we have our nine point circle for our triangle ABC that passes through the three mid-points of the sides, the three feet of the altitudes, and the three mid-points of the segments from the respective vertices to orthocenter.