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
at E
[I. 10]
,
and let BE be joined
and produced in a straight line
to F;
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
BE to
EF,
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;
[C.N. 5]
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, Elements, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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