(related to Proposition: Prop. 11.26: Construction of Solid Angle equal to Given Solid Angle)

- Let $AB$ be the given straight line, and $A$ the given point on it, and $D$ the given solid angle, contained by the rectilinear angles $EDC$, $EDF$, and $FDC$.
- So, it is necessary to construct a solid angle equal to the solid angle $D$ on the straight line $AB$, and at the point $A$ on it.

- For let some random point $F$ have been taken on $DF$, and let $FG$ have been drawn from $F$ perpendicular to the plane through $ED$ and $DC$ [Prop. 11.11], and let it meet the plane at$G$, and let $DG$ have been joined.
- And let $BAL$, equal to the angle $EDC$, and $BAK$, equal to $EDG$, have been constructed on the straight line $AB$ at the point $A$ on it [Prop. 1.23].
- And let $AK$ be made equal to $DG$.
- And let $KH$ have been set up at the point $K$ at right angles to the plane through $BAL$ [Prop. 11.12].
- And let $KH$ be made equal to $GF$.
- And let $HA$ have been joined.
- I say that the solid angle at $A$, contained by the (plane) angles $BAL$, $BAH$, and $HAL$, is equal to the solid angle at $D$, contained by the (plane) angles $EDC$, $EDF$, and $FDC$.
- For let $AB$ and $DE$ have been cut off (so as to be) equal, and let $HB$, $KB$, $FE$, and $GE$ have been joined.
- And since $FG$ is at "right angles to the reference plane":bookofproofs$2212 ($EDC$), it will also make right angles with all of the straight lines joined to it which are also in the reference plane [Def. 11.3] .
- Thus, the angles $FGD$ and $FGE$ are right angles.
- So, for the same (reasons), the angles $HKA$ and $HKB$ are also right angles.
- And since the two (straight lines) $KA$ and $AB$ are equal to the two (straight lines) $GD$ and $DE$, respectively, and they contain equal angles, the base $KB$ is thus equal to the base $GE$ [Prop. 1.4].
- And $KH$ is also equal to $GF$.
- And they contain right angles (with the respective bases).
- Thus, $HB$ (is) also equal to $FE$ [Prop. 1.4].
- Again, since the two (straight lines) $AK$ and $KH$ are equal to the two (straight lines) $DG$ and $GF$ (respectively), and they contain right angles, the base $AH$ is thus equal to the base $FD$ [Prop. 1.4].
- And $AB$ (is) also equal to $DE$.
- So, the two (straight lines) $HA$ and $AB$ are equal to the two (straight lines) $DF$ and $DE$ (respectively).
- And the base $HB$ (is) equal to the base $FE$.
- Thus, the angle $BAH$ is equal to the angle $EDF$ [Prop. 1.8].
- So, for the same (reasons), $HAL$ is also equal to $FDC$.
- And $BAL$ is also equal to $EDC$.
- Thus, (a solid angle) has been constructed, equal to the given solid angle at $D$, on the given straight line $AB$, at the given point $A$ on it.
- (Which is) the very thing it was required to do.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"