PROPOSITION 5.
If two triangles have their sides proportional,
the triangles will be equiangular and will have those angles equal which the corresponding sides subtend.
Let
ABC,
DEF be two triangles having their sides proportional, so that,
as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as
BA is to
AC, so is
ED to
DF; I say that the triangle
ABC is equiangular with the triangle
DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle
ABC to the angle
DEF, the angle
BCA to the angle
EFD, and further the angle
BAC to the angle
EDF.
For on the straight line
EF, and at the points
E,
F on it, let there be constructed the angle
FEG equal to the angle
ABC, and the angle
EFG equal to the angle
ACB; [
I. 23]
therefore the remaining angle at A is equal to the remaining angle at G. [I. 32]
Therefore the triangle
ABC is equiangular with the triangle
GEF.
Therefore in the triangles
ABC,
GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [
VI. 4]
therefore, as AB is to BC, so is GE to EF.
But, as
AB is to
BC, so by hypothesis is
DE to
EF;
therefore, as DE is to EF, so is GE to EF. [V. 11]
Therefore each of the straight lines
DE,
GE has the same ratio to
EF;
therefore DE is equal to GE. [V. 9]
For the same reason
DF is also equal to GF.
Since then
DE is equal to
EG, and
EF is common,
the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle
DEF is equal to the angle
GEF, [
I. 8] and the triangle
DEF is equal to the triangle
GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
DFE is also equal to the angle
GFE,
and the angle EDF to the angle EGF.
And, since the angle
FED is equal to the angle
GEF, while the angle
GEF is equal to the angle
ABC, therefore the angle
ABC is also equal to the angle
DEF.
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.
