Proof
Let be cut at ; consider the rectangle on , . By
Proposition II.1, taking as the uncut line and the two segments
, of itself as the cut line, the rectangle on and
equals the rectangle on and together with the
rectangle on and — but the rectangle on and is
the square on by Definition II.1. Re-expressing the rectangle
on and via II.1 again as the rectangle on , plus
the rectangle on , (which is the square on , again by
Definition II.1) and combining, we obtain:
the rectangle on and equals the rectangle on and
plus the square on . By Common Notion 2 the equality is
preserved when rearranged.
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Full neighborhood
Depends on (3)
- II.1Proposition II.1If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by…
- II.1Definition II.1Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle.
- 2Common notion 2If equals be added to equals, the wholes are equal.
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