Given three segments whose lengths are 1, a, and b, construct segments of length a + b, |a – b|, ab, $\displaystyle \frac{a}{b}$, and $\displaystyle \sqrt{ab}$. These are Euclidean constructions.
Hello, ReneePatt!
Surely, you can do the first two . . .
Given three segments whose lengths are: $\displaystyle 1,\;a,\text{ and }b,$
construct segments of lengths:
$\displaystyle (1)\;a + b \qquad (2)\;|a - b| \qquad (3)\;ab \qquad (4)\;\frac{a}{b} \qquad (5)\;\sqrt{ab}$
$\displaystyle (3)\;ab$
On a horizontal line, measure off: .$\displaystyle OP = 1,\;PQ = a$
Code:o - - - o - - - o - - - O 1 P a Q
Through $\displaystyle O$ draw a diagonal line.
On the diagonal, measure off $\displaystyle OR = b$
Draw $\displaystyle PR.$
Code:* * * R * o b * / * / * / o - - - o - - - o - - - O 1 P a Q
Through $\displaystyle Q$ construct a line parallel to $\displaystyle PR$, cutting the diagonal at $\displaystyle S.$
Code:S o x * / * / R * / o / b * / / * / / * / / o - - - o - - - o - - - O 1 P a Q
Then: .$\displaystyle x \,=\,RS \,=\,ab$
Proof
From similar triangles: .$\displaystyle \frac{b+x}{1+a} \:=\:\frac{b}{1}$
And we have: .$\displaystyle b + x \:=\:b + ab \quad\Rightarrow\quad x \:=\:ab$
1. Draw a line with the length (a + b)
2. This line is the diameter of a circle.
3. Construct a perpendicular line at the end of a = begin of b. This line intersect the circle line. Construct a right triangle with (a + b) as hypotenuse .
4. According to Euclid's theorem you have in a right triangle:
$\displaystyle a \cdot b = h^2~\implies~\boxed{h = \sqrt{a \cdot b}}$
Hello, Renee!
$\displaystyle (4)\;\frac{a}{b}$
On a horizontal line, measure off: .$\displaystyle OP = b,\;PQ = 1$
Code:o - - - o - - - o - - - O b P 1 Q
Through $\displaystyle O$ draw a diagonal line.
On the diagonal, measure off $\displaystyle OR = a$
Draw $\displaystyle PR.$
Code:* * * R * o a * / * / * / o - - - o - - - o - - - O b P 1 Q
Through $\displaystyle Q$ construct a line parallel to $\displaystyle PR$, cutting the diagonal at $\displaystyle S.$
Code:S o x * / * / R * / o / a * / / * / / * / / o - - - o - - - o - - - O b P 1 Q
Then: .$\displaystyle x \,=\,RS \,=\,\frac{a}{b}$
Proof
From similar triangles: .$\displaystyle \frac{a+x}{b+1} \:=\:\frac{a}{b}$
And we have: .$\displaystyle ab+bx \:=\:ab +a \quad\Rightarrow\quad bx \:=\:a \quad\Rightarrow\quad x \:=\:\frac{a}{b}$