In a triangle the greater angle has the longer side opposite to it.

Explanation

Given: Let `ABC` be a triangle in which the `angleABC` is greater than the `angleACB`.
To prove: `AC>AB`
Proof:Let us try to prove this by contradiction. let us assume that `AC` is not longer than `AB`.Then two cases will arise:
Case I : `AC=AB`
Case II : `AC< AB`
If `AC = AB`, `triangleABC` would has been isosceles triangle and `angleABC=angleBAC`
This contradicts to the given condition.
In the II case: The side `BC` would have been longer than `AC` and consequently the `angleABCBut from the theorem "If in a triangle two sides are unequal, then the angle opposite to longer side is greater than the angle opposite to shorter side".
Again this contradicts the given condition.
Thus the only remaining possibility is that the side `AC` is longer than the side `BC`.
Thus `AC>AB`.
Hence Proved