Construction of Solid Angle equal to Given Solid Angle

Proof

 * Euclid-XI-26.png

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 $DF$.

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

Let $DG$ be joined.

Using :
 * 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 $AK = DG$.

Using :
 * 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 :
 * $FG$ 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 :
 * $KB = GE$

But also:
 * $KH = GF$

Therefore from :
 * $HB = FE$

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

From :
 * $AH = FD$

But also:
 * $AB = DE$

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

We also have that $HB = FE$.

Therefore from :
 * $\angle BAH = \angle EDF$

For the same reason:
 * $\angle HAL = \angle FDC$

and:
 * $\angle BAL = \angle EDC$

Hence the result.