### Square and Square Root Construction by Compass and Straightedge

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.

Page created  .