# M2_8_8_5_7

Author:
angela

## 8_5_7

Existence of the Nagel point. Let triangle ABC be a triangle, let TA be the point at which the A-excircle is tangent to BC, let TB be the point at which the B-excircle is tangent to AC, and let TC be the point at which the C-excircle is tangent to AB. Prove that the segments ATA, BTB , and C TC are concurrent. Hint: Use Exercise 4.5.2 to prove that the ratios in this problem are the reciprocals of the ratios in Exercise 8.5.6.

## Solution:

I did not complete proof 4.5.2 and thus did not use it in this proof. However, I did use Ceva's Theorem. First, let TC denote the excircle of ΔABC opposite the vertex C. TC is tangent to the side AB and the side lines AC and BC in points F, T and U. Since, CT and CU are the two tangents from C to TC, they are equal in length (1) CT = CU. For a similar reason (2) AT=AF. (3) BU = BF. (1)-(3) immediately imply that F is a perimeter splitter, which means AC + AF = BC + BF. If a, b, c are the side lengths of the ΔABC and p its semiperimeter, then we can say b + AF = a + BF = p, so that AF = p - b BF = p - a. Similarly, we obtain additional four identities: AE = p - c, CE = p - a, CD = p - b, BD = p - c. Ceva's Theorem verified: AE/CE·CD/BD·BF/AF= (p-c)/(p-a)·(p-b)/(p-c)·(p-a)/(p-b) = 1.