-Prakash Pant

There are several amazing results in Mathematics of which I believe one of them is Nine Point Circle. As the name suggests, this amazing Nine point circle passes through nine well known points in the triangle viz 3 midpoints of sides (M1,M2,M3), 3 legs of altitudes (P1,P2,P3) and 3 mid points of line joining vertex to the orthocenter (E1,E2,E3). 

The man who first recognized that the Circumcircle of Orthic triangle (P1P2P3) also cuts the original triangle ABC at its midpoints was Karl Wilhelm Feurbach who first named it a six-point circle. Later, Olry Terquem  proved that this circle also passes through Euler Points (midpoint of line joining vertex to Orthocentre), thus giving it the name Nine Point Circle as well as Feurbach’s Circle, Terquem’s Circle, Euler’s Circle, MedioScribed Circle and many more.

Now,  the amazing thing that we will discuss today is that “The Center of the Nine Point Circle is the midpoint of Euler’s Line.”  And that too, in a very simple and elegant way using Homothety. If you have not heard about Euler’s line, it is simply the line joining Circumcentre and Orthocentre of a triangle. It gained its reputation as,  for any triangle, it always passes through the centroid of the triangle whose proof is  left to readers. Next,  Homethety is just the geometric transformation using an enlargement  factor and an enlargement center. 

Here in the figure given below,  We have ∆ABC with its circumcircle. Three altitudes of triangle AD, BE and CF  intersect at Orthocentre H. These altitudes when extended intersect the circumcircle at D’, E’ and F’ respectively. J, K, L are the midpoints of the three sides of the triangle which are connected with the Orthocentre H. HJ, HK and HL intersect the circumcircle at J’,K’ and L’ respectively. 

Now, since we know that the reflection of Orthocentre lies on the circumcircle of the triangle, HD=DD’, HE=EE’ and HF=FF’. Then, reflect H across L and call it L2. Now, HL=LL2 and BL=LC. Then , BHCH2 is a parallelogram. Hence, <BH2C = <BHC = 180-<HBC-<HCB  = (90-<HBC)+(90-<HCB)= <C+<B = 180-A. Here,as  180-<A = BH2C, thus  A,B, H2,C are concyclic and then L’=L2.So, HL=LL’. Similarly, HJ=JJ’ and HK=KK’.

Finally, to conclude our discussion, all we need to do is to consider the homothety  h(H,1/2) centered at H that takes the circumcircle and maps it to the nine point circle. Well, it would immediately be implied that E’ would be mapped onto E, F’ would be mapped onto F, D’ would be mapped onto D as the distances of Orthocentre from former points are twice the distances of the Orthocentre from the later distances. For eg: D’H is twice of DH, E’H is twice of EH, F’H is twice of FH and so on. And in the similar way, J’ would be mapped onto J, K’ would be mapped onto K and L’ would be mapped onto L as a result of this homothety. This homothety with a factor of ½ will shrink our circumcircle into the Nine-Point Circle. 

But, wait here is an additional benefit of that. It actually also gives us a clear view of the location of the center of the Nine Point Circle. Obviously when you think about it, the distance from the circumcentre to the Orthocentre, as any other point, will also be reduced by ½. And that clearly implies that, in the Euler line between Circumcentre  and Orthocentre, our center of Nine Point Circle moves halfway from the Circumcentre to Orthocentre i.e. at its midpoint. So we clearly showed that the Centre of the Nine Point Circle is the midpoint of the Euler’s Line with an easy Visualization through Homothety.

Now, with this very idea, what I want you to do is to prove that the radius of the Nine Point Circle is R/2 where R is the circumradius of the triangle ABC.  Enjoy proving…..