# M2_8_8_5_7

- Author:
- angela

## GeoGebra Construction:

## 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 T

_{C}denote the excircle of ΔABC opposite the vertex C. T_{C}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 T_{C}, 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.## New Resources

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