| In any triangle the sum of any two sides is greater than the remaining one. | ||
| Let ABC be a triangle.
I say that in the triangle ABC the sum of any two sides is greater than the remaining one, that is, the sum of BA and AC is greater than BC, the sum of AB and BC is greater than AC, and the sum of BC and CA is greater than AB. | ||
| Draw BA through to the point D, and make DA equal to CA. Join DC. | Post.2
I.3 Post.1 | |
| Since DA equals AC, therefore the angle ADC also equals the angle ACD. Therefore the angle BCD is greater than the angle ADC. | I.5
C.N.5 | |
| Since DCB is a triangle having the angle BCD greater than the angle BDC, and the side opposite the greater angle is greater, therefore DB is greater than BC. | I.19 | |
| But DA equals AC, therefore the sum of BA and AC is greater than BC.
Similarly we can prove that the sum of AB and BC is also greater than CA, and the sum of BC and CA is greater than AB. | ||
| Therefore in any triangle the sum of any two sides is greater than the remaining one. | ||
| Q.E.D. | ||
