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

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