Euclid Book 6 Proposition 8
Statement
If in a right triangle (△ACB) a perpendicular () is drawn from the right angle (∠ACB) to the base (), then the triangles (△ACD, △BCD) adjoining the perpendicular are similar both to the whole and to one another.
CD
AB
Computational Explanation
GeometricScene[{{A.,B.,C.,D.},{}},{Triangle[{A.,C.,B.}],PlanarAngle[{A.,C.,B.}]90°,GeometricAssertion[{Line[{{A.,D.,B.},{C.,D.}}]},"Perpendicular"]},{GeometricAssertion[{Triangle[{A.,C.,D.}],Triangle[{C.,B.,D.}],Triangle[{A.,B.,C.}]},"Similar"]}]
GeometricAssertion[{Triangle[{A.,C.,D.}],Triangle[{C.,B.,D.}],Triangle[{A.,B.,C.}]},Similar] |
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Explanations
Let be a right-angled triangle having the angle right, and let be drawn from perpendicular to ; I say that each of the triangles , is similar to the whole and, further, they are similar to one another.
For, since the angle is equal to the angle , for each is right, and the angle at is common to the two triangles and , therefore the remaining angle is equal to the remaining angle ;therefore the triangle is equiangular with the triangle .
Therefore, as which subtends the right angle in the triangle is to which subtends the right angle in the triangle , so is itself which subtends the angle at in the triangle to which subtends the equal angle in the triangle , and so also is to which subtends the angle at common to the two triangles.
Therefore the triangle is both equiangular to the triangle and has the sides about the equal angles proportional.
Therefore the triangle is similar to the triangle .
Similarly we can prove that the triangle is also similar to the triangle ; therefore each of the triangles , is similar to the whole .
I say next that the triangles, are also similar to one another.
For, since the right angle is equal to the right angle , and moreover the angle was also proved equal to the angle at , therefore the remaining angle at is also equal to the remaining angle ;therefore the triangle is equiangular with the triangle .
Therefore, as which subtends the angle in the triangle is to which subtends the angle at in the triangle equal to the angle , so is itself which subtends the angle at in the triangle to which subtends the angle in the triangle equal to the angle at , and so also is to , these sides subtending the right angles; therefore the triangle is similar to the triangle .
ABC
BAC
AD
A
BC
ABD
ADC
ABC
For, since the angle
BAC
ADB
B
ABC
ABD
ACB
BAD
ABC
ABD
Therefore, as
BC
ABC
BA
ABD
AB
C
ABC
BD
BAD
ABD
AC
AD
B
Therefore the triangle
ABC
ABD
Therefore the triangle
ABC
ABD
Similarly we can prove that the triangle
ABC
ADC
ABD
ADC
ABC
I say next that the triangles
ABD
ADC
For, since the right angle
BDA
ADC
BAD
C
B
DAC
ABD
ADC
Therefore, as
BD
BAD
ABD
DA
C
ADC
BAD
AD
B
ABD
DC
DAC
ADC
B
BA
AC
ABD
ADC