Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle : it is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D.
Make the parallelogram BEFG equal to the triangle
C, and having the angle EBG equal to the angle D, so
that BE may be in the same straight line with AB ; [I. 42.
produce FG to H ;
through A draw AH parallel to BG or EF, [I. 31.
and join HB. [I. 31.
Then, because the straight line HF falls on the parallels
AH, EF, the angles AHF, HFE are together equal to two right angles. [I. 29.
Therefore the angles BHF, HFE are together less than two right angles.
But straight lines which with another straight line make the interior angles on the same side together less than two right angles will meet on that side, if produced far enough. [Ax. 12.
Therefore HB and FE will meet if produced ;
let them meet at K.
Through K draw KL parallel to EA or FH ; [I. 31.
and produce HA, GB to the points L, M.
Then HLKF is a parallelogram, of which the diameter
is HK; and AG, ME are parallelograms about HK; and
LB, BF are the complements.
Therefore LB is equal to BF. [I. 43.
But BF is equal to the triangle C [Construction.</br.>
Therefore LB is equal to the triangle C [Axiom 1.
