Theorem : if in a triangle two angles are not equal, then the side opposite to the greater angle is longer than the side opposite to the smaller angle.
given: let ABC be a triangle in which the angle ABC is greater than the angle ACB.
TPT: AC > AB
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 angle ABC = angle BAC
this contradicts to the given condition.
in the II nd case: the side BC would has been longer than AC and consequently the. angle ABC < angle BAC.
but from the theorem " If in a triangle two sides are un equal , 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