PROPOSITION 6.
If two triangles have one angle equal to one angle and the sides about the equal angles proportional,
the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.
Let
ABC,
DEF be two triangles having one angle
BAC equal to one angle
EDF and the sides about the equal angles proportional, so that,
as BA is to AC, so is ED to DF; I say that the triangle
ABC is equiangular with the triangle
DEF, and will have the angle
ABC equal to the angle
DEF, and the angle
ACB to the angle
DFE.
For on the straight line
DF, and at the points
D,
F on it, let there be constructed the angle
FDG equal to either of the angles
BAC,
EDF, and the angle
DFG equal to the angle
ACB; [
I. 23]
therefore the remaining angle at B is equal to the remaining angle at G. [I. 32]
Therefore the triangle
ABC is equiangular with the triangle
DGF.
Therefore, proportionally, as
BA is to
AC, so is
GD to
DF. [
VI. 4]
But, by hypothesis, as
BA is to
AC, so also is
ED to
DF; therefore also, as
ED is to
DF, so is
GD to
DF. [
V. 11]
Therefore
ED is equal to
DG; [
V. 9] and
DF is common;
therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
DFG is equal to the angle
DFE,
and the angle DGF to the angle DEF.
But the angle
DFG is equal to the angle
ACB; therefore the angle
ACB is also equal to the angle
DFE.
And, by hypothesis, the angle
BAC is also equal to the angle
EDF; therefore the remaining angle at
B is also equal to the remaining angle at
E; [
I. 32]
therefore the triangle ABC is equiangular with the triangle DEF.
Therefore etc. Q. E. D.
