Therefore the angles ACD and ADC are equal.

Also the angle BCD is greater than the angle ACD.
(CN-5)

Let ABC be the given triangle.

Assume that side BC is greater than side AB and AC.

We will show this is impossible.

The other cases are proved similarly.

 

Extend the side AB to D where AD is the same as AC.

 

 

Construct the line DC.

The triangle CAD is isosceles.

 

 

If BC is greater than AB, AD (that is, BD) and CD, it follows that the angle BDC is the greatest angle of the triangle BDC.
(Prop I-18)

This is impossible because the angle BCD is greater. Thus BC cannot be greater than both CD and BD.

Assume that BC is greater than BD but less than CD.

Then angle ABC is the greater angle of the triangle BCD. In particular the angle ABC is greater than the angle BCA.
(CN-5)

 

(CN-x)

(Prop I-xx)

(Post-xx)

Extend the side AC to E where AE is the same as AB. So, BD equals CE.

Arguing as before that BC is greater than CE, we conclude that angle BCA is greater than angle ABC.

It is impossible that angle ABC be both greater than and less than angle BCA. Therefore, the side BC is not greater than the sum of the sides AC and AB.

QED