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