Let AB be the given straight line, C the given triangle and D the given rectilineal angle;
thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42] ; let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF [I. 31]

Let HB be joined.

Then, since the straight line HF falls upon the parallels AH, EF,
the angles AHF, HFE are equal to two right angles. [I. 29]
Therefore the angles BHG, GFE are less than two right angles;
and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]
therefore HB, FE, when produced, will meet.

Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31]
and let HA, GB be produced to the points L, M.

Then HLKF is a parallelogram,
HK is its diameter, and AG, ME are parallelograms, and LB, BF the so-called complements, about HK;
therefore LB is equal to BF. [I. 43]

But BF is equal to the triangle C;
therefore LB is also equal to C. [C.N. 1]

And, since the angle GBE is equal to the angle ABM, [I. 15]
while the angle GBE is equal to D,
the angle ABM is also equal to the angle D.

Therefore 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. Q.E.F.