# Part 1

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#### Rational Trigonometry: ** Spread in a circle **** Part 1**

**Spread in a circle**

**– Part 2 – Part 3 – Part 4 – Part 5 – Part 6 –**

*Spread *measurement in a circle** **

Following the definition of *spread* as the ratio of an opposite *quadrance* to the hypotenuse *quadrance* in a right-angled triangle, consider (following Euclid’s theorem on intersecting chords) forming two **similar** right-angled triangles in the semicircles **ACB** and **AC’B**. Let the diameter **AB** equal 1.

Then, comparing corresponding sides (noting **OC** = **OC**’ by symmetry)** **

x/y (in **AOC’**) = y/(1 – x) (in **COB**) and therefore y² = x (1 – x).

But since (in **COB**) h² = y² + (1 – x)² = x (1 – x) + (1 – x) (1 – x) = (1) (1 – x)

the spread at vertex **B**, s(**B**) = y² / h² = x (1 – x) / (1 – x) = x (*)

This only measures spread when one of the lines is a **diameter** of the circle however, so what about a spread between two** oblique **lines?

Beginning with chords **A’B** and **C’B**, if we rotate **A’B **onto the diameter **AB **this takes **C’B **onto chord **CB**. We can see that chords/arcs **A’A **and **C’C **need to be equal.

The perpendicular to **C’B **through **A’**, which meets the circle also at **C**, is parallel to chord **AC’**, also perpendicular to **C’B **(a right angle subtended by a diameter). Parallel chords, with midpoints collinear with the circle centre, are bisected by a common radius which bisects the associated inner and outer arcs **A’AC’C **and **AC’** leaving arcs **A’A **and **C’C **equal by symmetry, and making **A’AC’ **and **AC’C **equal, This allows spread to again be measured by dropping a perpendicular from **C** to diameter **AB** at **O**.

It thus appears a circle provides direct connection with the *geometric *measurement of spread. However, it is not the circle that determines this construction but the property of perpendicularity. A circle serves simply as a device for constructing these perpendiculars, suggesting that spread could instead be constructed *without *an auxiliary circle. Is this possible?

Let us re-examine our basic figure by adding further parallel segments:

Here a second perpendicular, this time to chord **A’B **through point **C’**, has been drawn. As expected by symmetry this meets the circle, at point **D**, on the *same *perpendicular line through **C** and **O**.

We observe that new segments, **C’G **and **A’G**, are parallel and equal to existing segments **AA’ **and **AC’ **respectively. Since **AA’ **was equal by symmetry to **C’C**, **C’G **equals **C’C**. Since **A’C **is perpendicular to **C’B**, triangles **C’GE **and **C’CE **are congruent, making segments **GE **and **CE **equal, and **E** (where perpendicular **A’C **meets **C’B**) the midpoint of **GC**. Likewise, point **F** (on **A’B**) is midpoint to **GD**. Finally, **O** is necessarily the midpoint of chord **CD**.

Putting this together, triangle **OEF **will be *medial *to triangle **CDG**, ensuring that perpendicular *segments* from **A** to lines **C’B **or **A’B** followed by parallel segments **C’F** or **A’E** followed by parallel segments from **F** or **E** meet on the diameter at the *same *position as the previously dropped perpendicular from **C** or **D** on the circle.

This construction measures the spread of oblique lines without the need for an auxiliary circle. We amend our basic figure accordingly:

Here we leave chord **CD **of the previous construction visible for reference. The only remaining objects necessary to obtain a measurement of *spread *are:

- a pair of oblique
*lines*through point**B**(whose spread is to be measured) - a reference point
**A**(defining the*unit segment***AB**) - a pair of
*perpendiculars***A**to the given pair of*lines*through**B** - a
*segment**parallel*to the*alternate line*from**A***perpendicular to the alternate line*through**B** - a second
*segment**parallel*to the*original line*from**A**meeting**AB**at position**O**.

**The spread at B is then the ratio of segment AO to AB.**

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