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Given the unit length , and the segment
of length
construct
.
Solution: Let be the right angle triangle with
and
. Let the axis of the side
meets the line
in point
. Construct the semicircle
with center in
and radius
. If
is the second point of the diameter passing through points
and
, then
.
Given the unit length , and the segment
of length
construct
.
Solution: Let be semicircle with diameter
, where
and
. Let the perpendicular on
at
meets the semicircle
in point
. Then
.
This construction is a special case of Euclid’s solution of Proposition 14 in Book II of his Elements: To construct a square equal to a given rectilineal figure.
Due to Proposition 45 of Book I, the solution can be reduced to the case when the given rectilineal figure is a rectangle .
The proof (not the original Euclid’s one) follows easily from two later propositions:
Proposition 35 (Book III): If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.
Proposition 3 (Book III): If in a circle a straight line through the center bisect a straight line not through thecenter, it also cuts it at right angles; and if it cut it at right angles it also bisects it.
Cite this web-page as:
Štefan Porubský: Square and Square Root Construction by Compass and Straightedge.