Triangle inequality and Chapter V review

Theorem: any side of a triangle is smaller than the sum of the other two sides: AC < AB + BC. Proof: on the extension of AB lay off BD = BC; △BCD is isosceles, so ∠BCD = ∠BDC. Ray CB passes inside ∠ACD, so ∠ACD > ∠BCD = ∠ADC; in △ACD the larger side lies opposite the larger angle, hence AC < AD = AB + BC. Corollary 1: for any points A, B, C not on one line, AC < AB + BC, AB < AC + BC and BC < AB + AC. Corollary 2: any side of a triangle is greater than the difference of the other two sides. So if two sides are a and b, the third side x satisfies |a − b| < x < a + b. A triangle with sides 3, 4, 8 does not exist because 3 + 4 < 8. Such an example can also serve as a counterexample.