Euclid Book 6 Proposition 32
Statement
If two triangles (△ABC, △CDE) which have two sides of one proportional to two sides of the other (), and the contained angles (∠ABC, ∠CDE) equal, are joined at a point (C), so as to have their corresponding sides parallel, the remaining sides are in the same line.
AB
BC
CD
DE
Computational Explanation
GeometricScene{{A.,B.,C.,D.,E.},{}},GeometricAssertion[{Triangle[{A.,B.,C.}],Triangle[{C.,D.,E.}]},"Counterclockwise"],,PlanarAngle[{A.,B.,C.}]PlanarAngle[{C.,D.,E.}],GeometricAssertion[{Line[{A.,B.}],Line[{C.,D.}]},"Parallel"],GeometricAssertion[{Line[{B.,C.}],Line[{D.,E.}]},"Parallel"],{GeometricAssertion[{A.,C.,E.},"Collinear"]}
EuclideanDistance[A.,B.]
EuclideanDistance[B.,C.]
EuclideanDistance[C.,D.]
EuclideanDistance[D.,E.]
GeometricAssertion[{A.,C.,E.},Collinear] |
 |
Explanations
Let , be two triangles having the two sides , proportional to the two sides , , so that, as is to , so is to , and parallel to , and to ; I say that is in a straight line with . For, since is parallel to , and the straight line has fallen upon them, the alternate angles , are equal to one another.
For the same reason the angle is also equal to the angle ; so that the angle is equal to the angle .
And, since, are two triangles having one angle, the angle at A, equal to one angle, the angle at , and the sides about the equal angles proportional, so that, as is to , so is to , therefore the triangle is equiangular with the triangle ;therefore the angle is equal to the angle .
But the angle was also proved equal to the angle ; therefore the whole angle is equal to the two angles , .
Let the angle be added to each; therefore the angles , are equal to the angles , , .
But the angles, , are equal to two right angles; therefore the angles , are also equal to two right angles.
Therefore with a straight line, and at the point on it, the two straight lines , not lying on the same side make the adjacent angles , equal to two right angles; therefore is in a straight line with .
ABC
DCE
BA
AC
DC
DE
AB
AC
DC
DE
AB
DC
AC
DE
BC
CE
AB
DC
AC
BAC
ACD
For the same reason the angle
CDE
ACD
BAC
CDE
And, since
ABC
DCE
D
BA
AC
CD
DE
ABC
DCE
ABC
DCE
But the angle
ACD
BAC
ACE
ABC
BAC
Let the angle
ACB
ACE
ACB
BAC
ACB
CBA
But the angles
BAC
ABC
ACB
ACE
ACB
Therefore with a straight line
AC
C
BC
CE
ACE
ACB
BC
CE