# Construction of Solid Angle from Three Plane Angles any Two of which are Greater than Other Angle/Lemma

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## Theorem

In the words of Euclid:

*But how it is possible to take the square on $OR$ equal to that area by which the square on $AB$ is greater than the square on $LO$, we can show as follows.*

(*The Elements*: Book $\text{XI}$: Proposition $23$ : Lemma)

## Proof

Let the straight lines $AB$ and $LO$ be set out.

Le $AB > LO$.

Let the semicircle $ACB$ be described on $AB$.

Using Proposition $1$ of Book $\text{IV} $: Fitting Chord Into Circle:

- Let $AC = LO$ be fitted into the semicircle $ACB$.

Let $BC$ be joined.

From Proposition $31$ of Book $\text{III} $: Relative Sizes of Angles in Segments:

- $\angle ACB$ is a right angle.

Therefore from Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem:

- $AB^2 = AC^2 + CB^2$

Hence:

- $AB^2 - AC^2 = CB^2$

But $AC = LO$.

Therefore:

- $AB^2 - LO^2 = CB^2$

So if we then cut of $OR = BC$:

- $AB^2 - LO^2 = OR^2$

$\blacksquare$

## Historical Note

This proof is Proposition $23$ of Book $\text{XI}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions