PROPOSITION 4.
In equiangular triangles the sides about the equal angles are proportional,
and those are corresponding sides which subtend the equal angles.
Let
ABC,
DCE be equiangular triangles having the angle
ABC equal to the angle
DCE, the angle
BAC to the angle
CDE, and further the angle
ACB to the angle
CED; I say that in the triangles
ABC,
DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
For let
BC be placed in a straight line with
CE.
Then, since the angles
ABC,
ACB are less than two right angles, [
I. 17] and the angle
ACB is equal to the angle
DEC, therefore the angles
ABC,
DEC are less than two right angles; therefore
BA,
ED, when produced, will meet. [
I. Post. 5]
Let them be produced and meet at
F.
Now, since the angle
DCE is equal to the angle
ABC,
BF is parallel to CD. [I. 28]
Again, since the angle
ACB is equal to the angle
DEC,
AC is parallel to FE. [I. 28]
Therefore
FACD is a parallelogram; therefore
FA is equal to
DC, and
AC to
FD. [
I. 34]
And, since
AC has been drawn parallel to
FE, one side of the triangle
FBE, therefore, as
BA is to
AF, so is
BC to
CE. [
VI. 2]
But
AF is equal to
CD;
therefore, as BA is to CD, so is BC to CE, and alternately, as
AB is to
BC, so is
DC to
CE. [
V. 16]
Again, since
CD is parallel to
BF, therefore, as
BC is to
CE, so is
FD to
DE. [
VI. 2]
But
FD is equal to
AC;
therefore, as BC is to CE, so is AC to DE, and alternately, as
BC is to
CA, so is
CE to
ED. [
V. 16]
Since then it was proved that,
as AB is to BC, so is DC to CE, and, as
BC is to
CA, so is
CE to
ED; therefore,
ex aequali, as
BA is to
AC, so is
CD to
DE. [
V. 22]
Therefore etc. Q. E. D.
