# Proof: By Euclid • 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.

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