To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let $AB$ be the given straight line, $C$ the given triangle and $D$ the given rectilineal angle;thus it is required to apply to the given straight line $AB$, in an angle equal to the angle $D$, a parallelogram equal to the given triangle $C$.Let the parallelogram $BEFG$ be constructed equal to the triangle $C$, in the angle $EBG$ which is equal to $D$ [Prop. 1.42]; let it be placed so that $BE$ is in a straight line with $AB$; let $FG$ be drawn through to $H$, and let $AH$ be drawn through $A$ parallel to either $BG$ or $EF$. [Prop. 1.31]Let $HB$ be joined.Then, since the straight line $HF$ falls upon the parallels $AH$, $EF$,the angles $AHF$, $HFE$ are equal to two right angles. [Prop. 1.29]Therefore the angles $BHG$, $GFE$ are less than two right angles;and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]therefore $HB$, $FE$ when produced, will meet.Let them be produced and meet at $K$; through the point $K$ let $KL$ be drawn parallel to either $EA$ or $FH$, [Prop. 1.31]and let $HA$, $GB$ be produced to the points $L$, $M$.Then $HLKF$ is a parallelogram,$HK$ is its diameter, and $AG$, $ME$ are parallelograms, and $LB$, $BF$ the so-called complements, about $HK$;therefore $LB$ is equal to $BF$. [Prop. 1.43]therefore $LB$ is also equal to $C$. [C.N. 1]And, since the angle $GBE$ is equal to the angle $ABM$, [Prop. 1.15]while the angle $GBE$ is equal to $D$,the angle $ABM$ is also equal to the angle $D$.Therefore the parallelogram $LB$ equal to the given triangle $C$ has been applied to the given straight line $AB$, in the angle $ABM$ which is equal to $D$.Q.E.F.