PROPOSITION 7.
If two triangles have one angle equal to one angle,
the sides about other angles proportional,
and the remaining angles either both less or both not less than a right angle,
the triangles will be equiangular and will have those angles equal,
the sides about which are proportional.
Let
ABC,
DEF be two triangles having one angle equal to one angle, the angle
BAC to the angle
EDF, the sides about other angles
ABC,
DEF proportional, so that, as
AB is to
BC, so is
DE to
EF, and, first, each of the remaining angles at
C,
F less than a right angle; I say that the triangle
ABC is equiangular with the triangle
DEF, the angle
ABC will be equal to the angle
DEF, and the remaining angle, namely the angle at
C, equal to the remaining angle, the angle at
F.
For, if the angle
ABC is unequal to the angle
DEF, one of them is greater.
Let the angle
ABC be greater; and on the straight line
AB, and at the point
B on it, let the angle
ABG be constructed equal to the angle
DEF. [
I. 23]
Then, since the angle
A is equal to
D, and the angle
ABG to the angle
DEF, therefore the remaining angle
AGB is equal to the remaining angle
DFE. [
I. 32]
Therefore the triangle
ABG is equiangular with the triangle
DEF.
Therefore, as
AB is to
BG, so is
DE to
EF [
VI. 4]
But, as
DE is to
EF, so by hypothesis is
AB to
BC; therefore
AB has the same ratio to each of the straight lines
BC,
BG; [
V. 11]
therefore BC is equal to BG, [V. 9] so that the angle at
C is also equal to the angle
BGC. [
I. 5]
But, by hypothesis, the angle at
C is less than a right angle; therefore the angle
BGC is also less than a right angle; so that the angle
AGB adjacent to it is greater than a right angle. [
I. 13]
And it was proved equal to the angle at
F; therefore the angle at
F is also greater than a right angle.
But it is by hypothesis less than a right angle : which is absurd.
Therefore the angle
ABC is not unequal to the angle
DEF; therefore it is equal to it.
But the angle at
A is also equal to the angle at
D; therefore the remaining angle at
C is equal to the remaining angle at
F. [
I. 32]
Therefore the triangle
ABC is equiangular with the triangle
DEF.
But, again, let each of the angles at
C,
F be supposed not less than a right angle; I say again that, in this case too, the triangle
ABC is equiangular with the triangle
DEF.
For, with the same construction, we can prove similarly that
BC is equal to BG; so that the angle at
C is also equal to the angle
BGC. [
I. 5]
But the angle at
C is not less than a right angle; therefore neither is the angle
BGC less than a right angle.
Thus in the triangle
BGC the two angles are not less than two right angles: which is impossible. [
I. 17]
Therefore, once more, the angle
ABC is not unequal to the angle
DEF; therefore it is equal to it.
But the angle at
A is also equal to the angle at
D; therefore the remaining angle at
C is equal to the remaining angle at
F. [
I. 32]
Therefore the triangle
ABC is equiangular with the triangle
DEF.
Therefore etc. Q. E. D.
