◀ ▲ ▶Branches / Geometry / Elements-euclid / Book--6-similar-figures / Proof: By Euclid

Proof: By Euclid

(related to Proposition: 6.08: Perpendicular in Right-Angled Triangle makes two Similar Triangles)

• For since (angle) $BAC$ is equal to $ADB$ - for each (are) right angles - and the (angle) at $B$ (is) common to the two triangles $ABC$ and $ABD$, the remaining (angle) $ACB$ is thus equal to the remaining (angle) $BAD$ [Prop. 1.32].
• Thus, triangle $ABC$ is equiangular to triangle $ABD$.
• Thus, as $BC$, subtending the right angle in triangle $ABC$, is to $BA$, subtending the right angle in triangle $ABD$, so the same $AB$, subtending the angle at $C$ in triangle $ABC$, (is) to $BD$, subtending the equal (angle) $BAD$ in triangle $ABD$, and, further, (so is) $AC$ to $AD$, (both) subtending the angle at $B$ common to the two triangles [Prop. 6.4].
• Thus, triangle $ABC$ is equiangular to triangle $ABD$, and has the sides about the equal angles proportional.
• Thus, triangle $ABC$ [is] [similar]bookofproofs$1983 to triangle $ABD$ [Def. 6.1] .
• So, similarly, we can show that triangle $ABC$ is also similar to triangle $ADC$.
• Thus, [triangles] $ABD$ and $ADC$ are each similar to the whole (triangle) $ABC$.
• So I say that triangles $ABD$ and $ADC$ are also similar to one another.
• For since the right angle $BDA$ is equal to the right angle $ADC$, and, indeed, (angle) $BAD$ was also shown (to be) equal to the (angle) at $C$, thus the remaining (angle) at $B$ is also equal to the remaining (angle) $DAC$ [Prop. 1.32].
• Thus, triangle $ABD$ is equiangular to triangle $ADC$.
• Thus, as $BD$, subtending (angle) $BAD$ in triangle $ABD$, is to $DA$, subtending the (angle) at $C$ in triangle $ADC$, (which is) equal to (angle) $BAD$, so (is) the same $AD$, subtending the angle at $B$ in triangle $ABD$, to $DC$, subtending (angle) $DAC$ in triangle $ADC$, (which is) equal to the (angle) at $B$, and, further, (so is) $BA$ to $AC$, (each) subtending right angles [Prop. 6.4].
• Thus, triangle $ABD$ is similar to triangle $ADC$ [Def. 6.1] .
• Thus, if, in a right-angled triangle, a (straight line) is drawn from the right angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another.
• (Which is) the very thing it was required to show.
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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"