Euclid Book 6 Proposition 6
Statement
If two triangles (△ABC, △DEF) have a pair of corresponding angles equal (∠BAC∠EDF), and the ratios of the sides adjacent to the corresponding angles are equal (), then those triangles have equal corresponding angles.
AB
AC
DE
DF
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.,B.,C.}]PlanarAngle[{D.,E.,F.}],PlanarAngle[{A.,C.,B.}]PlanarAngle[{D.,F.,E.}]}
EuclideanDistance[A.,B.]
EuclideanDistance[A.,C.]
EuclideanDistance[D.,E.]
EuclideanDistance[D.,F.]
PlanarAngle[{A.,B.,C.}]PlanarAngle[{D.,E.,F.}] |
 |
Explanations
Let , be two triangles having one angle equal to one angle and the sides about the equal angles proportional, so that, as is to , so is to ; I say that the triangle is equiangular with the triangle , and will have the angle equal to the angle , and the angle to the angle .
For on the straight line, and at the points , F on it, let there be constructed the angle equal to either of the angles , , and the angle equal to the angle ;therefore the remaining angle at is equal to the remaining angle at .
Therefore the triangle is equiangular with the triangle . Therefore, proportionally, as is to , so is to .
But, by hypothesis, as is to , so also is to ; therefore also, as is to , so is to .
Therefore is equal to ;and is common; therefore the two sides , are equal to the two sides , ; and the angle is equal to the angle ; therefore the base is equal to the base , and the triangle is equal to the triangle , and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend.
Therefore the angle is equal to the angle , and the angle to the angle .
But the angle is equal to the angle ; therefore the angle is also equal to the angle .
And, by hypothesis, the angle is also equal to the angle ; therefore the remaining angle at is also equal to the remaining angle at ;therefore the triangle is equiangular with the triangle .
ABC
DEF
BAC
EDF
BA
AC
ED
DF
ABC
DEF
ABC
DEF
ACB
DFE
For on the straight line
DF
D
FDG
BAC
EDF
DFG
ACB
B
G
Therefore the triangle
ABC
DGF
BA
AC
GD
DF
But, by hypothesis, as
BA
AC
ED
DF
ED
DF
GD
DF
Therefore
ED
DG
DF
ED
DF
GD
DF
EDF
GDF
EF
GF
DEF
DGF
Therefore the angle
DFG
DFE
DGF
DEF
But the angle
DFG
ACB
ACB
DFE
And, by hypothesis, the angle
BAC
EDF
B
E
ABC
DEF