To apply a parallelogram equal to a given triangle to a given straight-line in a given rectilinear angle. Let $AB$ be the given straight-line,  $C$ the given triangle, and $D$ the given rectilinear angle. So it is required to apply a parallelogram equal to the given triangle $C$ to the given straight-line $AB$ in an angle equal to (angle) $D$.  Let the parallelogram $BEFG$, equal to the triangle $C$, have been constructed in the angle $EBG$, which is equal to $D$ [Prop.~1.42]. And let it have been placed so that $BE$ is straight-on to $AB$.  And let $FG$ have been drawn through to $H$, and let $AH$ have been drawn through A parallel to either of $BG$ or $EF$ [Prop.~1.31], and let $HB$ have been joined. And since the straight-line $HF$ falls across the parallels $AH$ and $EF$, the (sum of the) angles $AHF$ and $HFE$ is thus equal to two right-angles [Prop.~1.29]. Thus, (the sum of) $BHG$ and $GFE$ is less than two right-angles. And (straight-lines) produced to infinity from (internal angles whose sum is) less than two right-angles meet together [Post.~5]. Thus, being produced, $HB$ and $FE$ will meet together. Let them have been produced, and let them meet together at $K$. And let $KL$ have been drawn through point $K$ parallel to either of $EA$ or $FH$ [Prop.~1.31]. And let $HA$ and $GB$ have been produced to points $L$ and $M$ (respectively). Thus, $HLKF$ is a parallelogram, and $HK$ its diagonal. And $AG$ and $ME$ (are) parallelograms, and $LB$ and $BF$ the so-called complements, about $HK$. Thus, $LB$ is equal to $BF$ [Prop.~1.43]. But, $BF$ is equal to triangle $C$. Thus, $LB$ is also equal to $C$. Also, since angle $GBE$ is equal to $ABM$ [Prop.~1.15], but $GBE$ is equal to $D$, $ABM$ is thus also equal to angle $D$.  Thus, 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 $D$. (Which is) the very thing it was required to do.