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) $D$. <prf> 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 $D$.