In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.
Let ABC be a triangle, and let
one side of it BC be
produced to D;
I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.
Let AC be bisected
and let BE be joined
and produced in a straight line
let EF be made equal to BE [I. 3] , let FC be joined [Post. 1] , and let AC be drawn through to G. [Post. 2]
Then, since AE is equal
to EC, and
the two sides AE, EB are equal to the two sides CE, EF respectively;
and the angle AEB is equal to the angle FEC,
for they are vertical angles. [I. 15]
Therefore the base AB is equal to the base FC,
and the triangle ABE is equal to the triangle CFE,
and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]
therefore the angle BAE is equal to the angle ECF.
But the angle ECD is greater
than the angle ECF;
therefore the angle ACD is greater than the angle BAE.
Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15] , can be proved greater than the angle ABC as well.
Therefore etc. Q.E.D.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
Next: Proposition 17
Previous: Proposition 15
This proposition in other editions: