Euclid Book 6 Proposition 7b
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 non-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 each of the angles at, be supposed not less than a right angle; I say that the triangle is equiangular with the triangle .
For, with the same construction, we can prove similarly that is equal to ; so that the angle at is also equal to the angle .
But the angle at is not less than a right angle; therefore neither is the angle less than a right angle.
Thus in the triangle the two angles are not less than two right angles: which is impossible.
Therefore, once more, 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 each of the angles at
C
F
ABC
DEF
For, with the same construction, we can prove similarly that
BC
BG
C
BGC
But the angle at
C
BGC
Thus in the triangle
BGC
Therefore, once more, the angle
ABC
DEF
But the angle at
A
D
C
F
Therefore the triangle
ABC
DEF