Let [math]\bigtriangleup[/math] ABC be any triangle. Let the circle centered at D be the excircle for the point C. Let the circle centered at E be the excircle of B and the circle centered at F be the excircle of A. Notice that the circle centered a D is tangent at L, P, J and the circle centered at E is tangent at G, H, I and the circle centered at F is tangent at M, N, K. [br]Claim the Cevians through the points P, H, and N are concurrent. Since CL and CJ are tangent to the circle centered a D then for the circle centered at C orthogonal to the circle centered at D and then CL and CJ are radii of this circle. Thus CL is congruent to CJ. The same argument [br]hold for AK congruent to AM and BI congruent to BG. Now consider the circle centered at B that is orthogonal to the circle centered at D. A similar argument yields the congruency of BL and BP. If we apply the same idea to the circle centered at C and orthogonal to the circle [br]centered at E we see that CH and CG are congruent. More importantly BL, [br]BP, CH, CG are all congruent.We can construct congruences between AJ, AP, CN and CK. As well as BM, BN, AL and AH. Notice that AP[math]\cong[/math]CN and AH[math]\cong[/math]BN and CH[math]\cong[/math]BP. Thus AH[math]\times[/math]CN[math]\times[/math]BP = BN[math]\times[/math]AP[math]\times[/math]CH Thus [br](AH/CH)[math]\times[/math](CN/BN)[math]\times[/math](BP/AP) = 1[br]Therefore by Ceva's theorem the segments AN, BH and CP are concurrent at the Nagel point.

The relationship has already been explored in homework #3 problem 6 and problem 22. Problem 6 gave us that a tangent line to a circle is perpendicular to a unique radius. Problem 22 gave us that two orthogonal circles intersect in only two points. This allows me to add the red circle centered at A and the blue circles centered at B and C. This construction allows to find these halfway points with more than the point tangent to the excircle.