Proportion and Thales' theorem

Four segments $\displaystyle a$, $\displaystyle b$, $\displaystyle c$, $\displaystyle d$ are proportional when the ratio of the first two equals the ratio of the last two:

From one proportion you get others – each by a step of algebra:

• multiplying out: $\displaystyle {ad}={bc}$;
• swapping the middle terms: $\displaystyle {\frac{a}{c}}={\frac{b}{d}}$;
• adding $\displaystyle 1$ to both sides: $\displaystyle {\frac{a+b}{b}}={\frac{c+d}{d}}$;
• subtracting $\displaystyle 1$ from both sides: $\displaystyle {\frac{a-b}{b}}={\frac{c-d}{d}}$;
• turning both sides upside down after adding: $\displaystyle {\frac{a}{a+b}}={\frac{c}{c+d}}$.

Example: the point $\displaystyle D$ divides the segment $\displaystyle AB$ so that $\displaystyle {\frac{AD}{DB}}={\frac{2}{3}}$. Adding $\displaystyle 1$ to both sides: $\displaystyle {\frac{AD+DB}{DB}}={\frac{5}{3}}$, that is $\displaystyle {\frac{AB}{DB}}={\frac{5}{3}}$. So $\displaystyle DB$ is $\displaystyle \frac{3}{5}$ of $\displaystyle AB$, and $\displaystyle AD$ is $\displaystyle \frac{2}{5}$ of it.

The point $\displaystyle D$ is on the segment $\displaystyle AB$, and $\displaystyle {AB}={20}$.
$\displaystyle {\frac{AD}{DB}}={\frac{2}{3}}$.
Find $\displaystyle AD$.
(A hint: $\displaystyle {AD}={x}$, and then $\displaystyle {DB}={20-x}$.)

Theorem: If parallel lines cut equal segments on one arm of an angle, they cut equal segments on the other arm too.

Given: $\displaystyle D$, $\displaystyle F$, $\displaystyle H$ are on one arm of the angle $\displaystyle A$, and $\displaystyle {AD}={DF}={FH}$; $\displaystyle DE\parallel FG\parallel HI$, with $\displaystyle E$, $\displaystyle G$, $\displaystyle I$ on the other arm.
To prove: $\displaystyle {AE}={EG}={GI}$.

The proof below uses two theorems of the midsegment lesson: in the triangle $\displaystyle AFG$, the line $\displaystyle DE$ goes through the midpoint of $\displaystyle AF$ parallel to $\displaystyle FG$; in the trapezoid $\displaystyle DEIH$, the line $\displaystyle FG$ goes through the midpoint of the leg $\displaystyle DH$ parallel to the bases.

In the same way the theorem holds for any number of equal segments, one trapezoid after another.

Thales’ theorem: Two parallel lines that cut the arms of an angle cut proportional segments on them.

In a triangle: if $\displaystyle D$ is on the side $\displaystyle AB$, $\displaystyle E$ is on the side $\displaystyle AC$, and $\displaystyle DE\parallel BC$, then

The proof below is for a ratio of whole numbers, here $\displaystyle {\frac{AD}{DB}}={\frac{2}{3}}$; any other ratio of whole numbers goes the same way.

• $\displaystyle AD$ is $\displaystyle 2$ equal parts and $\displaystyle DB$ is $\displaystyle 3$ parts of the same length: the points $\displaystyle K$, $\displaystyle D$, $\displaystyle L$, $\displaystyle M$ divide $\displaystyle AB$ into $\displaystyle 5$ equal parts;
• through each of them draw a line parallel to $\displaystyle BC$, to the side $\displaystyle AC$ – at $\displaystyle N$, $\displaystyle E$, $\displaystyle P$, $\displaystyle Q$;
• the parallel lines cut equal segments on $\displaystyle AB$, so by the theorem above they cut equal segments on $\displaystyle AC$ too;
• so $\displaystyle AE$ is $\displaystyle 2$ of these parts and $\displaystyle EC$ is $\displaystyle 3$ of them: $\displaystyle {\frac{AE}{EC}}={\frac{2}{3}}={\frac{AD}{DB}}$.

The theorem holds for every ratio, not only for a ratio of whole numbers; the proof of that is beyond this course.

The properties of a proportion turn $\displaystyle {\frac{AD}{DB}}={\frac{AE}{EC}}$ into two more – a part to the whole side:

and $\displaystyle {\frac{DB}{AB}}={\frac{EC}{AC}}$.

Example: $\displaystyle DE\parallel BC$, $\displaystyle {AD}={4}$, $\displaystyle {DB}={6}$ and $\displaystyle {AE}={5}$. By Thales’ theorem $\displaystyle {\frac{4}{6}}={\frac{5}{EC}}$, so $\displaystyle {4\cdot EC}={30}$ and $\displaystyle {EC}={7.5}$.

In the triangle $\displaystyle ABC$, $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$.
$\displaystyle DE\parallel BC$.
$\displaystyle {AD}={6}$, $\displaystyle {AB}={15}$ and $\displaystyle {AC}={20}$.
Find $\displaystyle AE$.

With a ruler and a compass you can divide a segment $\displaystyle AB$ into $\displaystyle 5$ equal parts:

• draw a ray from $\displaystyle A$, at any angle to $\displaystyle AB$;
• with the compass, mark $\displaystyle 5$ equal segments on the ray, one after another, from $\displaystyle A$ to a point $\displaystyle T$;
• join $\displaystyle T$ to $\displaystyle B$;
• through each of the other marks draw a line parallel to $\displaystyle TB$.

These parallel lines cut equal segments on the ray, so they cut equal segments on $\displaystyle AB$ too – $\displaystyle 5$ equal parts.

To divide $\displaystyle AB$ in the ratio $\displaystyle 2:3$, draw only the parallel through the second mark: it cuts $\displaystyle AB$ at a point $\displaystyle D$ with $\displaystyle {\frac{AD}{DB}}={\frac{2}{3}}$, by Thales’ theorem.

Describe how to divide a segment $\displaystyle AB$ in the ratio $\displaystyle 3:4$ with a ruler and a compass.
Explain why the point you find divides $\displaystyle AB$ in that ratio.

In the triangle $\displaystyle ABC$, $\displaystyle D$ is on the side $\displaystyle AB$ and $\displaystyle E$ is on the side $\displaystyle AC$.
$\displaystyle DE\parallel BC$.
Prove: $\displaystyle {AD\cdot EC}={DB\cdot AE}$.