Euclid Book 6 Proposition 7a
Statement
If two triangles (△ABC, △DEF) have a pair of corresponding angles equal (∠BAC∠EDF), the ratios of the sides adjacent to another pair of corresponding angles are equal (), and the remaining pair of angles are both acute, then those triangles are similar.
BA
BC
ED
EF
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.,F.},{}},Triangle[{{A.,B.,C.},{D.,E.,F.}}],PlanarAngle[{B.,A.,C.}]PlanarAngle[{E.,D.,F.}],PlanarAngle[{A.,C.,B.}]<90°,PlanarAngle[{D.,F.,E.}]<90°,,{GeometricAssertion[{Triangle[{A.,B.,C.}],Triangle[{D.,E.,F.}]},"Similar"]}
EuclideanDistance[B.,A.]
EuclideanDistance[B.,C.]
EuclideanDistance[E.,D.]
EuclideanDistance[E.,F.]
GeometricAssertion[{Triangle[{A.,B.,C.}],Triangle[{D.,E.,F.}]},Similar] |
 |
Explanations
Let , be two triangles having one angle equal to one angle, the angle to the angle , the sides about other angles , proportional, so that, as is to , so is to , and, first, each of the remaining angles at , less than a right angle; I say that the triangle is equiangular with the triangle , the angle will be equal to the angle , and the remaining angle, namely the angle at , equal to the remaining angle, the angle at .or, if the angle is unequal to the angle , one of them is greater.
Let the angle be greater; and on the straight line , and at the point on it, let the angle be constructed equal to the angle .
Then, since the angle is equal to , and the angle to the angle , therefore the remaining angle is equal to the remaining angle .
Therefore the triangle is equiangular with the triangle .
Therefore, as is to , so is to
But, as is to , so by hypothesis is to ; therefore has the same ratio to each of the straight lines , ;therefore is equal to ,so that the angle at is also equal to the angle .
But, by hypothesis, the angle at is less than a right angle; therefore the angle is also less than a right angle; so that the angle adjacent to it is greater than a right angle.
And it was proved equal to the angle at; therefore the angle at is also greater than a right angle.
But it is by hypothesis less than a right angle: which is absurd.
Therefore the angle is not unequal to the angle ; therefore it is equal to it.
But the angle at is also equal to the angle at ; therefore the remaining angle at is equal to the remaining angle at .
Therefore the triangle is equiangular with the triangle .
ABC
DEF
BAC
EDF
ABC
DEF
AB
BC
DE
EF
C
F
ABC
DEF
ABC
DEF
C
F
F
ABC
DEF
Let the angle
ABC
AB
B
ABG
DEF
Then, since the angle
A
D
ABG
DEF
AGB
DFE
Therefore the triangle
ABG
DEF
Therefore, as
AB
BG
DE
EF
But, as
DE
EF
AB
BC
AB
BC
BG
BC
BG
C
BGC
But, by hypothesis, the angle at
C
BGC
AGB
And it was proved equal to the angle at
F
F
But it is by hypothesis less than a right angle: which is absurd.
Therefore the angle
ABC
DEF
But the angle at
A
D
C
F
Therefore the triangle
ABC
DEF