Compass and straightedge construction of a square of an arbitrary line segment

Question

If I have some arbitrary length line $AB$ and a unit length line $CD$, how can I construct a line whose length is equal to the square of the length of line $AB$ using a compass and a straightedge?

Answer 1

It is assumed that the simple way to erect a normal at the end of a line segment by Ruler & Compass is already known.

Construct or place the line segments $CD,AP$ as perpendicular straight lines making points $D,A$ to coincide.

Join $CP$ and construct its perpendicular line $PB$ cutting extended line $CD$ at $B$.

Construct a circle $PCQB$ with $CB$ as diameter, mark the the center point of $CB$ as $O,$ center of circle for reference.

By constant product of line segments in a Circle $$ CD\cdot AB = AP^2.$$ From this theorem we obtain the straight line segment required as $AB$

since

$$AP=AQ=c, \,CD=1,\,$$

and so it follows

$$ AB=c^2.$$

Answer 2

Here's the more general construction for the product of two given lengths. On the sides of an angle of vertex $O$ construct points $A$ and $B$ such that $OA$ and $OB$ are the segments you want to multiply. On side $OB$ construct point $U$ such that $OU=1$. Let then $C$ be the intersection of side $OA$ with the line through $B$, parallel to $AU$. The similarity of triangles $OUA$ and $OBC$ gives then $$ OA:OU=OC:OB, \quad\text{that is:}\quad OC=OA\cdot OB. $$

Answer 3

Here's yet another construction, in the spirit of Narasimham's answer but with the generality of Aretino's, in that it allows you to construct a segment whose length is the product of any two segments:

Suppose you have two segments of length $a$ and $b$, and a unit-length segment. Choose any point $P$ and draw segments $AP$, $BP$, and $CP$ so that $A,P,B$ are collinear with $P$ between $A$ and $B$, and such that $|AP|=a$, $|BP|=b$, and $|CP|=1$. (See figure.)

Now, construct the circumcircle of triangle $ABC$. This can be done by constructing the perpendicular bisectors of any two of the triangle's sides; they meet at the circumcenter. (See figure.)

Now extend $CP$ until it reaches the other side of the circle at point $D$. Then $CP\cdot DP = AP \cdot BP$ implies that $DP = ab$. (See figure.)

Narasimham's answer is just a special case of this in which $a=b$ and in which the initial segments $AB$ and $PC$ are constructed to be perpendicular, neither of which is really necessary to make the argument work.