The angle BDA is exterior to the triangle BDC. Thus, the angle BDA is greater than the interior angle BCD.
(Prop I-16)

But the angle BDA is the same as angle ABD.

 

Let ABC be the given triangle with AC the greater side.

Cut off the segment on AC equal to AB.

Construct the segment BD.

The triangle ABD is isosceles.

 

Since angle ABD is less than angle CBA, it follows that angle DCB is less than angle CBA.
(CN-5)

Similarly, cutting of an amount equal to BC on AC, we prove that angle BAC is less than angle CBA.

Thus, the greater side subtends the greater angle.

QED

 

QED

(CN-x)

(Prop I-xx)

(Post-xx)