Euclid Book 6 Proposition 5
Statement
If two triangles (△ABC, △DEF) have equal corresponding ratios of adjacent sides (, , ), then they have equal corresponding angles.
AB
AC
DE
DF
CA
CB
FD
FE
AB
BC
DE
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.,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[C.,A.]
EuclideanDistance[C.,B.]
EuclideanDistance[F.,D.]
EuclideanDistance[F.,E.]
EuclideanDistance[A.,B.]
EuclideanDistance[B.,C.]
EuclideanDistance[D.,E.]
EuclideanDistance[E.,F.]
PlanarAngle[{B.,A.,C.}]PlanarAngle[{E.,D.,F.}] |
 |
Explanations
Let , be two triangles having their sides proportional, so that, as is to , so is to , as is to , so is to , and further, as is to , so is to ; I say that the triangle is equiangular with the triangle , and they will have those angles equal which the corresponding sides subtend, namely the angle to the angle , the angle to the angle , and further 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 the angle , 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 in the triangles, the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; therefore, as is to , so is to .
But, as is to , so by hypothesis is to ; therefore, as is to , so is to .
Therefore each of the straight lines, has the same ratio to ; therefore is equal to .
For the same reason is also equal to .
Since then is equal to , and is common, the two sides , are equal to the two sides , ; and the base is equal to the base ; therefore the angle is equal to the angle ,and the triangle is equal to the triangle , and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend.
Therefore the angle is also equal to the angle , and the angle to the angle .
And, since the angle is equal to the angle , while the angle is equal to the angle , therefore the angle is also equal to the angle .
For the same reason the angle is also equal to the angle , and further, the angle at to the angle at ; therefore the triangle is equiangular with the triangle .
ABC
DEF
AB
BC
DE
EF
BC
CA
EF
FD
BA
AC
ED
DF
ABC
DEF
ABC
DEF
BCA
EFD
BAC
EDF
For on the straight line
EF
E
FEG
ABC
EFG
ACB
A
G
Therefore the triangle
ABC
GEF
Therefore in the triangles
ABC
GEF
AB
BC
GE
EF
But, as
AB
BC
DE
EF
DE
EF
GE
EF
Therefore each of the straight lines
DE
GE
EF
DE
GE
For the same reason
DF
GF
Since then
DE
EG
EF
DE
EF
GE
EF
DF
FG
DEF
GEF
DEF
GEF
Therefore the angle
DFE
GFE
EDF
EGF
And, since the angle
FED
GEF
GEF
ABC
ABC
DEF
For the same reason the angle
ACB
DFE
A
D
ABC
DEF