the squares on the sides of a right triangle are equal to the square on the hypotenuse
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.. index:: proof, triangles

.. image:: elem.1.prop.47.a.png
   :align: right
   :width: 300px

.. image:: elem.1.prop.47.b.png
   :align: right
   :width: 300px

In right-angled triangles the square on [^I.47:1] the side subtending the right angle [^I.47:2] is equal to the squares on the sides containing the right angle.

===

Let `ABC` be a right-angled triangle having the angle `BAC` right;

I say that the square on `BC` is equal to the squares on `BA`, `AC`.

For let there be described on `BC` the square `BDEC`, and on `BA`, `AC` the squares `GB`, `HC`; [I.46] through `A` let `AL` be drawn parallel to either `BD` or `CE`, and let `AD`, `FC` be joined.

Then, since each of the angles `BAC`, `BAG` is right, it follows that with a straight line `BA`, and at the point `A` on it, the two straight lines `AC`, `AG` not lying on the same side make the adjacent angles equal to two right angles; 

- therefore `CA` is in a straight line with `AG`. [I.14]

For the same reason 

- `BA` is also in a straight line with `AH`.

And, since the angle `DBC` is equal to the angle `FBA`: for each is right: let the angle `ABC` be added to each; 

- therefore the whole angle `DBA` is equal to the whole angle `FBC`. [I.c.n.2]

And, since `DB` is equal to `BC`, and `FB` to `BA`, the two sides `AB`, `BD` are equal to the two sides `FB`, `BC` respectively, [^I.47:3]

- and the angle `ABD` is equal to the angle `FBC`; therefore the base `AD` is equal to the base `FC`, and the triangle `ABD` is equal to the triangle `FBC`. [I.4]

Now the parallelogram `BL` is double of the triangle `ABD`, for they have the same base `BD` and are in the same parallels `BD`, `AL`. [I.41]

And the square `GB` is double of the triangle `FBC`, for they again have the same base `FB` and are in the same parallels `FB`, `GC`. [I.41]

[But the doubles of equals are equal to one another.] [^I.47:4]

- Therefore the parallelogram `BL` is also equal to the square `GB`.

Similarly, if `AE`, `BK` be joined, the parallelogram `CL` can also be proved equal to the square `HC`; 

- therefore the whole square `BDEC` is equal to the two squares `GB`, `HC`. [I.c.n.2]

And the square `BDEC` is described on `BC`, 

- and the squares `GB`, `HC` on `BA`, `AC`.

Therefore the square on the side `BC` is equal to the squares on the sides `BA`, `AC`.

Therefore etc.

- Q. E. D.

## References

[I.4]: /elem.1.4 "Book 1 - Proposition 4"
[I.14]: /elem.1.14 "Book 1 - Proposition 14"
[I.41]: /elem.1.41 "Book 1 - Proposition 41"
[I.46]: /elem.1.46 "Book 1 - Proposition 46"
[I.c.n.2]: /elem.1.c.n.2 "Book 1 - Common Notion 2"

## Footnotes

[^I.47:1]: the square on
    , <foreign lang="greek">τὸ ἀπὸ...τετρἁγωνον</foreign>, the word <foreign lang="greek">ἀναγραφέν</foreign> or <foreign lang="greek">ἀναγεγραμμένον</foreign> being understood.

[^I.47:2]: subtending the right angle.
    Here <foreign lang="greek">ὑποτεινούσης</foreign>, <quote>subtending,</quote> is used with the simple accusative (<foreign lang="greek">τὴν ὀρθὴν γωνίαν</foreign>) instead of being followed by <foreign lang="greek">ὑπό</foreign> and the accusative, which seems to be the original and more orthodox construction. Cf. <a href="/elem.1.18">I. 18</a>, note.

      
[^I.47:3]: the two sides AB, BD....
    Euclid actually writes <quote>`DB`, `BA`,</quote> and therefore the equal sides in the two triangles are not mentioned in corresponding order, though he adheres to the words <foreign lang="greek">ἑκατέρα ἑκατέρα</foreign> <quote>respectively.</quote> Here `DB` is equal to `BC` and `BA` to `FB`.

[^I.47:4]: [But the doubles of equals are equal to one another.]
    Heiberg brackets these words as an interpolation, since it quotes a <title>Common Notion</title> which is itself interpolated. Cf. note on <a href="/elem.1.37">I. 37</a>, p. 332, and on interpolated <title>Common Notions</title>, pp. 223-4.