Euclid Book 6 Proposition 31
Statement
If similar polygons are drawn on each side of a right triangle (△ABC), then the area of the polygon (ABFG) on the hypotenuse () is equal to the sum of the areas of the other two polygons (ACHK, CBED).
AB
Computational Explanation
GeometricScene[{{A.,B.,C.,D.,E.,F.,G.,H.,K.},{}},{GeometricAssertion[{Triangle[{A.,B.,C.}]},"Counterclockwise"],PlanarAngle[{A.,C.,B.}]90°,GeometricAssertion[{Style[Polygon[{A.,K.,H.,C.}],Blue],Style[Polygon[{C.,D.,E.,B.}],Yellow],Style[Polygon[{B.,F.,G.,A.}],Green]},"Convex","Clockwise","Similar"]},{Area[Polygon[{A.,B.,F.,G.}]]Area[Polygon[{A.,C.,H.,K.}]]+Area[Polygon[{C.,B.,E.,D.}]]}]
Area[Polygon[{A.,B.,F.,G.}]]Area[Polygon[{A.,C.,H.,K.}]]+Area[Polygon[{C.,B.,E.,D.}]] |
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Explanations
Let be a right-angled triangle having the angle right; I say that the figure on is equal to the similar and similarly described figures on , .
Let be drawn perpendicular.
Then since, in the right-angled triangle, has been drawn from the right angle at perpendicular to the base , the triangles , adjoining the perpendicular are similar both to the whole and to one another.
And, since is similar to , therefore, as is to , so is to .
And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second.
Therefore, as is to , so is the figure on to the similar and similarly described figure on .
For the same reason also, as is to , so is the figure on to that on ; so that, in addition, as is to , , so is the figure on to the similar and similarly described figures on , .
But is equal to , ; therefore the figure on is also equal to the similar and similarly described figures on , .
ABC
BAC
BC
BA
AC
Let
AD
Then since, in the right-angled triangle
ABC
AD
A
BC
ABD
ADC
ABC
And, since
ABC
ABD
CB
BA
AB
BD
And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second.
Therefore, as
CB
BD
CB
BA
For the same reason also, as
BC
CD
BC
CA
BC
BD
DC
BC
BA
AC
But
BC
BD
DC
BC
BA
AC