PROPOSITION 44.

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

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.

C D H G F A B E L M K

Then, since the straight line HF falls[1] upon the parallels AH, EF,

[I. 29]the angles AHF, HFE are equal to two right angles.

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;

[I. 43]therefore LB is equal to BF.

But BF is equal to the triangle C;

[C. N. 1]therefore LB is also equal to C.

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.

See Commentary on Proposition I.44.