---
book: 6
number: 31
id: "VI.31"
kind: "theorem"
uses: ["[[book-6/proposition-8]]", "[[book-6/proposition-19]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.31"
license: "CC-BY-SA-4.0"
---

# VI.31

*In right*-*angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle*.

## Proof

Let *ABC* be a right-angled triangle having the angle *BAC* right; I say that the figure on *BC* is equal to the similar and similarly described figures on *BA*, *AC*.

Let *AD* be drawn perpendicular.

Then since, in the right-angled triangle *ABC*, *AD* has been drawn from the right angle at *A* perpendicular to the base *BC*, the triangles *ABD*, *ADC* adjoining the perpendicular are similar both to the whole *ABC* and to one another. [[book-6/proposition-8|VI. 8]]

And, since *ABC* is similar to *ABD*, therefore, as *CB* is to *BA*, so is *AB* to *BD*. [[book-6/definitions#Definition 1|VI. Def. 1]]

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. [[book-6/proposition-19|VI. 19, Por.]]

Therefore, as *CB* is to *BD*, so is the figure on *CB* to the similar and similarly described figure on *BA*.

For the same reason also, as *BC* is to *CD*, so is the figure on *BC* to that on *CA*; so that, in addition, as *BC* is to *BD*, *DC*, so is the figure on *BC* to the similar and similarly described figures on *BA*, *AC*.

But *BC* is equal to *BD*, *DC*; therefore the figure on *BC* is also equal to the similar and similarly described figures on *BA*, *AC*.

Therefore etc. Q. E. D.