#1) The three excircles of triangle ABC will be tangent to the (non-extended) sides at three points. Prove that the cevians joining the vertices of the triangle with these points of tangency are concurrent.

Let 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.
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
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
centered at E we see that CH and CG are congruent. More importantly BL,
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 APCN and AHBN and CHBP. Thus AHCNBP = BNAPCH Thus
(AH/CH)(CN/BN)(BP/AP) = 1
Therefore by Ceva's theorem the segments AN, BH and CP are concurrent at the Nagel point.

#1) b. The relationship between 46and 16

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.