Kitap 1
43
And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]
43
Therefore etc.
43
Q. E. D.
Proposition 44.
44
Enunciation To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.
44
Proof. 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.
44
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]
44
Let HB be joined.
44
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.
44
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.
44
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]
