Proof
Let be the given rectilineal figure. By Proposition I.45,
construct a parallelogram equal in area to , with the
parallelogram's angles right (apply I.46 on one of its sides if
necessary so it is a rectangle). If then is
already a square and the construction is complete. Suppose
; the case is symmetric.
Produce to , laying off on the produced line
(I.3). Bisect at (I.10). With centre and radius
describe the semicircle above . Produce
to meet the semicircle at .
Join ; then is a radius and equals . Apply II.5 to
bisected at and cut at : . By I.47 in the right triangle (right angle at
because ): . Subtract
from both expressions (Common Notion 3): .
But , so ; the square on equals
the rectangle , which equals the original figure .
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Full neighborhood
Depends on (9)
- I.3Proposition I.3Given two unequal straight lines, to cut off from the greater a straight line equal to the less.
- I.10Proposition I.10To bisect a given finite straight line.
- I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
- I.45Proposition I.45To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.
- I.46Proposition I.46On a given straight line to describe a square.
- I.47Proposition I.47In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides…
- II.5Proposition II.5If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole…
- 2Common notion 2If equals be added to equals, the wholes are equal.
- 3Common notion 3If equals be subtracted from equals, the remainders are equal.
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