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
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,
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,
and let HA, GB be produced to the points L, M.
Then HLKF is a
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
therefore LB is also equal to C. [C.N. 1]
And, since the angle GBE
is equal to the angle ABM,
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.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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