For the same reason
the angle ACB is also equal to the angle DFE, and further, the angle at
A to the angle at
D;
therefore the triangle ABC is equiangular with the triangle DEF.
Therefore etc. Q. E. D.
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]