Similarly, it is impossible that BC be greater than AC and BC.

We conclude that side AC is the greater of the three sides, which was to be proved.

QED

Let ABC be the given triangle, with angle CBA the greater angle.

We will show that AC is the greater side.

Suppose, instead that side AB is equal to or greater than AC.

If AB is equal to AC, then ABC is an isosceles triangle. Thus angles CBA and BCA are equal.
(Prop I-5)

Therefore angle CBA is not the greater, as was assumed.

 

Similarly BC is not equal to AC.

Now suppose that AB is greater than AC and BC. This means that AB is the greater side. Therefore angle BCA is the greater angle.
(Prop I-18) This is impossible as angle CBA is the greater.

 

 

 

 

 

(CN-x)

(Prop I-xx)

(Post-xx)