Intercept theorems

What is it about?

Two rays from a common starting point $Z$ are cut by two parallel lines: the first ray at $A$ and $A'$, the second at $B$ and $B'$. Then the intercept theorems hold.

1st intercept theorem: The segments on one ray are in the same ratio as the corresponding segments on the other ray:

2nd intercept theorem: The segments on the parallels are in the same ratio as the corresponding segments on a ray, measured from $Z$:

The “pieces” form ($\overline{ZA} : \overline{AA'}$) only works for the segments on the rays, not for the parallels.

The common value is the scale factor $k$. Triangle $ZA'B'$ is the image of $ZAB$ under an enlargement with centre $Z$: the two triangles are similar – same angles, every side times $k$. For $k > 1$ the figure is enlarged, for $0 < k < 1$ it is reduced, and for $k = 1$ the two parallels coincide.

If the parallels lie on different sides of $Z$, you get the X shape. The intercept theorems hold there in the same way; you always measure from $Z$. The simulation then shows a negative scale factor ($k < 0$: “other side of $Z$”); the ratios use lengths, i.e. $|k|$.

Without parallels it does not work: If $AB$ and $A'B'$ are not parallel, the ratios differ. Conversely, the converse of the 1st theorem holds: if $\overline{ZA'} : \overline{ZA} = \overline{ZB'} : \overline{ZB}$ (and $A'$ and $B'$ either both lie on the same side of $Z$ as $A$ and $B$, or both on the other side), then $AB \parallel A'B'$. The converse of the 2nd theorem is false: there can be a second point $B''$ with $\overline{A'B''} : \overline{AB} = k$ as well, although $A'B''$ is not parallel to $AB$.

Working with an unknown: Set up a proportion – best with the unknown $x$ in the numerator – and solve it by multiplying, e.g. $\frac{x}{\overline{ZB}} = \frac{\overline{ZA'}}{\overline{ZA}} \Rightarrow x = \frac{\overline{ZA'} \cdot \overline{ZB}}{\overline{ZA}}$.

Try it

1. Press “Slide the parallel”: the parallel slides from $AB$ to $A'B'$. Watch the coloured scale bars on the right – all three ratios count up together. What do the tick marks in the thick bars mean?
2. Drag $A'$ and turn the first ray with the handle at its outer end. Try $k = 0.5$, $k = 1$, a very acute angle and an obtuse angle. Do the ratios change?
3. X shape: press “Slide the parallel” – the parallel passes through $Z$ to the other side.
4. Compare the two “pieces” bars in the card “1st intercept theorem”: when do the dividing lines lie exactly on top of each other? Read in the card “2nd intercept theorem” why you must not use $\overline{AA'}$ there.
5. Not parallel: which bars no longer match? Press “Adjust B′ (converse)”. What happens to $A'B'$? What does the dashed circle around $A'$ with the point $B''$ show?
6. Unknown length x: choose different unknown lengths and let “Solve step by step” show you the working.
7. Tree height: play the measurement and then drag the sun. What happens at a sun elevation of 45°?
8. Pond width: drag $A'$ closer to $Z$. Why does $\overline{A'B'}$ get shorter, but the calculated width does not?

Tasks

Task 1: Missing length (1st theorem)

In a V shape with $AB \parallel A'B'$, $\overline{ZA} = 3\,\text{cm}$, $\overline{ZA'} = 6\,\text{cm}$ and $\overline{ZB} = 4\,\text{cm}$. Calculate $\overline{ZB'}$. Load task

Task 2: Working with pieces

You know $\overline{ZA} = 2.5\,\text{cm}$, $\overline{AA'} = 3.5\,\text{cm}$ and $\overline{ZB} = 3\,\text{cm}$ ($AB \parallel A'B'$, V shape). How long is $\overline{BB'}$? Load task

Task 3: The 2nd theorem

$\overline{ZA} = 4\,\text{cm}$, $\overline{ZA'} = 10\,\text{cm}$ and $\overline{AB} = 3\,\text{cm}$. Calculate $\overline{A'B'}$. Why would $\frac{\overline{A'B'}}{\overline{AB}} = \frac{\overline{AA'}}{\overline{ZA}}$ be wrong? Load task

Task 4: Unknown in the denominator

In the same figure, $\overline{ZB} = 5\,\text{cm}$, $\overline{ZB'} = 12.5\,\text{cm}$ and $\overline{A'B'} = 7.5\,\text{cm}$. Find $\overline{AB}$. Load task

Task 5: How tall is the tree?

A stick is $1.5\,\text{m}$ long. It is placed so that the tip of its shadow and the tip of the tree’s shadow meet. The stick stands $2\,\text{m}$ and the tree $16\,\text{m}$ from the shadow tip. Calculate the height of the tree. Load task

Task 6: Across the pond

To find the width $\overline{AB}$ of a pond, an X shape is marked out with $\overline{ZA'} = \frac14\,\overline{ZA}$ and $\overline{ZB'} = \frac14\,\overline{ZB}$. The measured length is $\overline{A'B'} = 6\,\text{m}$. How wide is the pond, and why is $A'B'$ automatically parallel to $AB$? Load task

Task 7: Parallel or not? (puzzle)

$\overline{ZA} = 3\,\text{cm}$, $\overline{ZA'} = 6\,\text{cm}$, $\overline{ZB} = 4\,\text{cm}$ and $\overline{ZB'} = 5.6\,\text{cm}$. Are $AB$ and $A'B'$ parallel? Where would $B'$ have to be? And: is it enough to check by measuring that $\overline{A'B'} = 2 \cdot \overline{AB}$?

Ideas for the classroom

• Use: About grade 9, similarity, enlargement and intercept theorems. The animation “Slide the parallel” can prompt a conjecture (“All ratios stay the same – but only if …”), and the cards then confirm it with numbers.
• Negative scale factors: Whether the X shape is described with $k < 0$ varies between classrooms – please check against the curriculum. That is why the slider works with the ratio of the lengths.
• Measuring outside: The tree and pond tasks can be recreated on the school grounds; “Play the measurement” shows the order of the steps. With “Show the result” switched off (show=0), the calculation mode and the applications become exercises.
• Common stumbling blocks:
  • In the 2nd theorem, $\overline{AA'}$ is used instead of $\overline{ZA'}$.
  • The theorems are applied without parallels.
  • In the X shape, lengths are added across $Z$.
  • Numerator and denominator are swapped between the two sides of the equation – the rule “unknown in the numerator, its partner below” helps here.
  • Parallelism is concluded from $\overline{A'B'} : \overline{AB} = k$ (the converse of the 2nd theorem is false).