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; letFG 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, ¹

• 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; I.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. I.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.

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Footnotes