To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.
[Desmos graphs can be found here]
Starting with a given rectilineal figure ABCD and a given angle E, we will construct a parallelogram equal to ABCD in an angle equal to E.
Join BD to express ABCD as two triangles: ABD and DBC.
Use Proposition 42 (construct a parallelogram in a given angle that is equal to a given triangle) to construct a parallelogram FGHK equal to triangle ABD in an angle FKH that is equal to angle E.
Use Proposition 44 (to construct a parallelogram to a given line in a given angle that is equal to a given triangle) to construct a parallelogram GHML on the line GH equal to triangle DBC in the angle GHM that is equal to angle E.
Since they are both equal to angle E, angles FKH and GHM are equal to one another.
Add angle KHG to both, so angles FKH and KHG together are equal to the angles GHM and KHG.
By Proposition 29 (a straight line falling on two parallel lines creates two interior angles on the same side that are together equal to two right angles) angles FKH and KHG together are equal to two right angles, so angles GHM and KHG are also together equal to two right angles.
So the straight line GH at a point H meets two straight lines KH and HM (that are not lying on the same side of GH) and makes two alternate angles equal to two right angles. By Proposition 14 (if with a straight line, and at a point on it, two straight lines not on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with each other), KH is in a straight line with HM.
The straight line GH falls upon two parallel lines KM and FG. By Proposition 29 (a straight line falling on two straight lines creates interior angles on the same side that together are equal to two right angles) the alternate angles HGF and GHM are equal to one another.
Add angle HGL to both, so the angles HGF and HGL together are equal to the angles GHM and HGL.
Again, by Proposition 29, the angles GHM and HGL together are equal to two right angles. And so the angles HGF and HGL are together equal to two right angles.
So, by Proposition 14, FG is in a straight line with GL (since GH meets the two lines that are not on the same side and makes two alternate angles equal to two right angles).
Since by Proposition 34 (opposite sides of a parallelogram are equal) FK is equal to GH, and GH is equal to LM, therefore FK is equal and parallel to LM by Proposition 30 (straight lines parallel to the same straight line are parallel to one another).
The lines FL and KM join the lines FK and LM at their extremities, so by Proposition 33 (the straight lines joining equal and parallel straight lines at the extremities which are in the same direction are themselves also equal and parallel) FL and KM are also equal and parallel.
Therefore FLMK is a parallelogram, and it is equal to the two parallelograms FGHK and GHML together.
Since FGHK is equal to triangle ABD and GHML is equal to the triangle DBC, the parallelogram FLMK is equal to the two triangles ABD and DBC and therefore is equal to the rectilineal figure ABCD.
Since FLMK is in the angle FKM and that angle is equal to angle E, we have constructed a parallelogram equal to the given figure in the given angle.
Whew – George