To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.
[Desmos graphs can be found here]
Let AB be the given straight line, C be the given triangle, and D the given angle.
We will apply a parallelogram to AB in an angle equal to D that is equal to the triangle C.
Use Proposition 42 (covers how to construct a parallelogram, in a given angle, that is equal to a given triangle) to construct the parallelogram BEFG that is equal to the triangle C in the angle EBG that is equal to D. Place the parallelogram so that BE is in a straight line with AB.
Let FG be drawn through to a point H such that AH is parallel to BG and EF. (Use Proposition 31: through a given point to draw a straight line that is parallel to a given line.)
Join HB.
Since the straight line HF falls on the parallel lines AH and EF, Proposition 29 (a straight line falling on two parallel lines creates two interior angles on the same side that are equal to two right angles) tells us that the angles AHF and HFE together are equal to two right angles.
Therefore the two angles BHG and GFE are less than two right angles.
(BHG is contained in and therefore less than AHF, and angle GFE is equal to HFE.)
Straight lines produced indefinitely from angles less than two right angles must meet, so the line through HB must meet the line through FE.
Let HB and FE be produced and meet at a point K.
Through K use Proposition 31 to draw a line KL that is parallel to FH and EA.
Produce HA to the point L and produce GB to a point M that lies on LK.
HLKF is a parallelogram, and HK is its diameter.
ABGH and MKEB are parallelograms contained in HLKF.
LMBA and BEFG are the complements about HK, and by Proposition 43 (in any parallelogram the complements of the parallelograms about the diameter are equal to one another) the two parallelograms LMBA and BEFG are equal.
Parallelogram BEFG was constructed to be equal to triangle C, so parallelogram LMBA is also equal to the triangle C.
Angle ABM is equal to angle GBE by Proposition 15 (if two straight lines intersect each other, they make the vertical angles equal to one another), and since angle GBE was constructed equal to angle D then the angle ABM is equal to angle of D.
Therefore parallelogram LMBA equal to the given triangle C has been applied to the given straight line AB, in the angle ABM equal to the given angle D.
The next proposition involves constructing a parallelogram in a given angle that is equal to a given rectilineal figure (instead of a given triangle) – George