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) $D$.
Given an arbitrary triangle (\(\triangle{C}\)), an arbitrary angle (\(\angle{D}\)), and an arbitrary segment (\(\overline{AB}\)), it is possible to construct a parallelogram (\(\boxdot{FGBE}\)) equal in area to the triangle $\triangle{C}$ which contains the given angle $\angle {D}=\angle{GBE}$ and has a side $\overline{GB}$ equal in length to the given segment $\overline{AB}$.
Proofs: 1