given: let ABC be a triangle in which the angle ABC is greater than the angle ACB.
TPT: 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. triangle ABC would has been isosceles triangle and
∠ABC=∠BAC
this contradicts to the given condition. in the II nd case: the side BC would has been longer than AC and consequently the
∠ABC<BAC
but from the theorem " If in a triangle two sides are unequal, then 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
