PROPOSITION 18.¶

In any triangle the greater side subtends the greater angle.

For let ABC be a triangle having the side AC greater than AB;

I say that the angle ABC is also greater than the angle BCA.

For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD be joined.

Then, since the angle ADB is an exterior angle of the triangle BCD,

[I. 16]it is greater than the interior and opposite angle DCB.

But the angle ADB is equal to the angle ABD,

since the side AB is equal to AD;

therefore the angle ABD is also greater than the angle ACB;

therefore the angle ABC is much greater than the angle ACB.

Therefore etc. Q. E. D.