Proof
Let the straight line be cut at random at the point .
Describe on the square (I.46), and through draw
parallel to either or (I.31), so that meets
at . The square is thereby divided into two rectangles:
contained by and (and since , this rectangle
is contained by and , by Definition II.1), and
contained by and (again, , so this rectangle is
contained by and ). Their sum (Common Notion 2) is the
whole square , which is the square on . Therefore the
rectangle on and together with the rectangle on and
equals the square on .
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Full neighborhood
Depends on (5)
- I.31Proposition I.31Through a given point to draw a straight line parallel to a given straight line.
- I.34Proposition I.34In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.
- I.46Proposition I.46On a given straight line to describe a square.
- 2Common notion 2If equals be added to equals, the wholes are equal.
- II.1Definition II.1Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle.
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