# Construction of Solid Angle equal to Given Solid Angle

## Theorem

In the words of Euclid:

*On a given straight line, and at a given point on it, to construct a solid angle equal to a given solid angle.*

(*The Elements*: Book $\text{XI}$: Proposition $26$)

## Proof

Let $AB$ be the given straight line.

Let $A$ be the given point on $AB$.

Let the angle at $D$, contained by the plane angles $\angle EDC, \angle EDF$ and $\angle FDC$, be the given solid angle.

Let an arbitrary point $F$ be taken on $DF$.

- Let $FG$ be drawn perpendicular to the plane through $ED$ and $DC$, meeting the plane at $G$.

Let $DG$ be joined.

Using Proposition $11$ of Book $\text{I} $: Construction of Equal Angle:

- Let $\angle BAL$ be constructed on the straight line $AB$ and at the point $A$ equal to $\angle EDC$, and $\angle BAK$ equal to $\angle EDG$.

Let $AK = DG$.

- Let $KH$ be set up from the point $K$ perpendicular to the plane through $BA$ and $AL$.

Let $KH = GF$.

Let $HA$ be joined.

It is to be demonstrated that the solid angle at $A$ contained by the plane angles $\angle BAL \angle BAH$ and $\angle HAL$ equals the angle at $D$, contained by the plane angles $\angle EDC, \angle EDF$ and $\angle FDC$.

Let $AB$ and $DE$ be cut off equal to each other.

Let $HB, KB, FE, GE$ be joined.

We have that $FG$ is perpendicular to the plane of reference.

So by Book $\text{XI}$ Definition $3$: Line at Right Angles to Plane:

- $FG$ is perpendicular to all straight lines which meet it and are in the plane of reference.

Therefore $\angle FGD$ and $\angle FGE$ are right angles.

For the same reason, $\angle HKA$ and $\angle HKB$ are right angles.

We have that $KA$ and $AB$ are equal to $GD$ and $DE$ respectively.

From Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

- $KB = GE$

But also:

- $KH = GF$

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

- $HB = FE$

We have that $AK$ and $KH$ are equal to $DG$ and $GF$ respectively.

From Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

- $AH = FD$

But also:

- $AB = DE$

Therefore $HA$ and $AB$ are equal to $DF$ and $DE$.

We also have that $HB = FE$.

Therefore from Proposition $8$ of Book $\text{I} $: Triangle Side-Side-Side Congruence:

- $\angle BAH = \angle EDF$

For the same reason:

- $\angle HAL = \angle FDC$

and:

- $\angle BAL = \angle EDC$

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $26$ 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