Euclid Book 6 Proposition 4
Statement
If two triangles (△ABC, △DEF) have equal corresponding angles, then the ratios of the sides adjacent to the corresponding angles are equal (, ,).
AB
AC
DE
DF
BA
BC
ED
EF
CA
CB
FD
FE
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.]
EuclideanDistance[B.,A.]
EuclideanDistance[B.,C.]
EuclideanDistance[E.,D.]
EuclideanDistance[E.,F.]
EuclideanDistance[C.,A.]
EuclideanDistance[C.,B.]
EuclideanDistance[F.,D.]
EuclideanDistance[F.,E.]
EuclideanDistance[A.,B.] EuclideanDistance[A.,C.] EuclideanDistance[D.,E.] EuclideanDistance[D.,F.] |
 |
Explanations
Let , be equiangular triangles having the angle equal to the angle , the angle to the angle , and further the angle to the angle ; I say that in the triangles , the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles.
For let be placed in a straight line with .
Then, since the angles, are less than two right angles, and the angle is equal to the angle , therefore the angles , are less than two right angles; therefore , , when produced, will meet.
Let them be produced and meet at.
Now, since the angle is equal to the angle , is parallel to .
Again, since the angle is equal to the angle , is parallel to .
Therefore is a parallelogram; therefore is equal to , and to .
And, since has been drawn parallel to , one side of the triangle , therefore, as is to , so is to .
But is equal to ; therefore, as is to , so is to , and alternately, as is to , so is to .
Again, since is parallel to , therefore, as is to , so is to .
But is equal to ; therefore, as is to , so is to , and alternately, as is to , so is to .
Since then it was proved that, as is to , so is to , and, as is to , so is to ; therefore, ex aequali, as is to , so is to .
ABC
DCE
ABC
DCE
BAC
CDE
ACB
CED
ABC
DCE
For let
BC
CE
Then, since the angles
ABC
ACB
ACB
DEC
ABC
DEC
BA
ED
Let them be produced and meet at
F
Now, since the angle
DCE
ABC
BF
CD
Again, since the angle
ACB
DEC
AC
FE
Therefore
FACD
FA
DC
AC
FD
And, since
AC
FE
FBE
BA
AF
BC
CE
But
AF
CD
BA
CD
BC
CE
AB
BC
DC
CE
Again, since
CD
BF
BC
CE
FD
DE
But
FD
AC
BC
CE
AC
DE
BC
CA
CE
ED
Since then it was proved that, as
AB
BC
DC
CE
BC
CA
CE
ED
BA
AC
CD
DE