How to interpret the ratio $\frac ab=\frac{p^2}{q^2}$ of line segments $a,b,p,q$ Let $a,b,p,q$ line segments (e.g. $a=3\text{cm}$ and so on). How can I interpret or construct the ratio $
\dfrac{a}{b}=\dfrac{p^2}{q^2}$ ? 
Is this something I can find on a suitable triangle or suitable right triangle?
 A: Choose an arbitrary segment ($\overline{P'Q'}$ in the figure), erect perpendiculars of $\overline{PP'}$ and $\overline{QQ'}$ of lengths $p$ and $q$, and construct $A$ and $B$ on $\overleftrightarrow{P'Q'}$ such that $\overline{AP}\perp\overline{PQ'}$ and $\overline{BQ}\perp\overline{P'Q}$.

Then
$$\left.\begin{align}
\frac{a}{p}=\frac{p}{c} \\[4pt]
\frac{b}{q}=\frac{q}{c}
\end{align}\right\}\quad\to\quad \frac{a}{b}=\frac{p^2/c}{q^2/c}=\frac{p^2}{q^2}$$
Here's a simple variant that constructs segment $\overline{AB}$ divided in the ratio $a:b = p^2:q^2$.

A: 
We give below two constructions that can be carried out to interpret the given ratio relationship.
$\bf{\underline{\text{Case}\space 1} :\space}$
Segaments $p$, $q$, and $a+b$ are given and we are looking for the segments $a$ and $b$, which satisfy the relationship
$$\frac{a}{b}= \frac{p^2}{q^2}. \tag{1}$$
We base our construction on Ceva’s Theorem. As shown in $\mathrm{Fig.\space 1}$, we start by drawing a line segment $AB$, the length of which is equal to $p+q$. Next, we need to draw a circle having the radius $p+q$ with the center at $A$. Then, we construct another circle having an arbitrary radius $r$ with the center at $B$, where $r \lt 2(p+q)$. The two circles intersect each other at two points. Select any one of these two points to complete the isosceles triangle $ABC$ with its apex at $A$. Mark the points $E$ and $F$ on $CA$ and $AB$ respectively such that $AE=q$ and $AF=p$. Join $BE$ and $CF$ to meet each other at point $G$. Now, construct the line segment $AG$ and extend it to cut $BC$ at $D$. A line segment $BM$, the length of which is equal to $a+b$,  is then drawn to make an arbitrary angle with $BC$. Finally, after joining the two points $C$ and $M$, a line parallel to $CM$ is drawn through $D$ to meet $BM$ at $N$.
When we apply Ceva’s theorem to the isosceles triangle $ABC$ shown in this configuration, we have,
$$\frac{c}{d}\times \frac{q}{p}\times \frac{q}{p}=1\qquad\rightarrow\qquad \frac{c}{d}=\frac{p^2}{q^2}. \tag{2}$$
Now, consider the $\triangle CBM$, where the line $DN$, which is parallel to the side $CM$, cut the other two sides $BC$ and $BM$. According to the Theorem 85 (Euclid VI. 2.), we have,
$$\frac{BN}{NM}=\frac{BD}{DC}\qquad\rightarrow\qquad \frac{a}{b}=\frac{c}{d}. \tag{3}$$
From (2) and (3), it follows that
$$\frac{a}{b}=\frac{p^2}{q^2}.$$

$\bf{\underline{\text{Case}\space 2} :\space}$ Segments $a$, $b$, and $p+q$ are given and we are looking for the segments $p$ and $q$, which satisfy the relationship
$$\frac{p^2}{q^2}=\frac{a}{b}. \tag{4}$$
The method described below with the help of $\mathrm{Fig.\space 2}$ is based on Pythagorean Theorem. We start by drawing the segments $AC$ and $CD$, such that $AC=a$ and $CD=b$. The point $O$ is the midpoint of $AD$. Construct the semicircle with its center at $O$ and its diameter equals to $OD$. When the perpendicular to $AD$ is drawn through $C$, it cuts the circumference of the semicircle at $E$. Now, as shown, mark $F$ on $AC$, such that $CF=CE$. A line segment $AB$, the length of which is equal to $p+q$, is then drawn to make an arbitrary angle with $AD$. Finally, after joining the two points $B$ and $F$, a line parallel to $BF$ is drawn through $C$ to meet $AB$ at $G$.
Applying Pythagorean Theorem to three right angle triangles, we obtain,
$$DE^2 + EA^2 =AD^2 = \left(a+b\right)^2 = a^2 +b^2 +2ab, \tag{$\triangle\space DEA$}$$
$$EA^2 =AC^2 + EC^2 = a^2 + EC^2, \tag{$\triangle\space ACE$}$$
$$DE^2 = EC^2 + CD^2 = EC^2 + b^2. \tag{$\triangle\space ECD$}$$
From these three equations, it follows that $\space EC^2 = ab$. Since $\space CF = EC\space$ by construction, we have $CF =  EC = \sqrt{ab}\space$ or, in otherworld, $\space AC : CF = \sqrt{a} : \sqrt{b}.\space$ Finally, using the Intercept Theorem, we shall write,
$$\frac{AG}{GB}=\frac{AC}{CF}=\frac{\sqrt{a}}{\sqrt{b}}\qquad\rightarrow\qquad \frac{AG^2}{GB^2}=\frac{p^2}{q^2}=\frac{a}{b}.$$
