Proof
Let and be the two straight lines, and let be cut at
random at the points , . Construct the rectangle contained by
and as follows. From draw at right angles to
(I.11) with equal to . Through draw parallel to
(I.31), and through , , in turn draw , ,
parallel to (I.31). Then the rectangle on the lines ,
is divided by the parallels , into the three rectangles
on , ; on , ; and on ,
(Definition II.1). By Common Notion 2, the whole rectangle equals
the sum of these parts: . The argument generalises to any number of cuts on .
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Full neighborhood
Depends on (5)
- I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
- 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.
- 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.
Required by (dependents) (3)
- II.3Proposition II.3If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the…
- II.7Proposition II.7If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to…
- II.8Proposition II.8If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together…
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