|In any triangle the side opposite the greater angle is greater.|
|Let ABC be a triangle having the angle ABC greater than the angle BCA.
I say that the side AC is greater than the side AB.
|If not, either AC equals AB or it is less than it.|
|Now AC does not equal AB, for then the angle ABC would equal the angle ACB, but it does not. Therefore AC does not equal AB.||I.5|
|Neither is AC less than AB, for then the angle ABC would be less than the angle ACB, but it is not. Therefore AC is not less than AB.||I.18|
|And it was proved that it is not equal either. Therefore AC is greater than AB.|
|Therefore in any triangle the side opposite the greater angle is greater.|
|Without going into details, the law of sines contains more precise information about the relation between angles and sides of a triangle than this and the last proposition did. The law of sines states that
Alternately, the first equation may be read a proportion
In other words, the sine of an angle in a triangle is proportional to the opposite side. (Proportions aren't defined in the Elements until Book V.)
Next proposition: I.20