Book 1
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.
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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]
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But BF is equal to the triangle C; therefore LB is also equal to C. [C.N. 1]
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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.
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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.
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Q. E. F.
Proposition 45.
45
Enunciation To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.
45
Proof. Let ABCD be the given rectilineal figure and E the given rectilineal angle; thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD.
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Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42] let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44]
45
Then, since the angle E is equal to each of the angles HKF, GHM, the angle HKF is also equal to the angle GHM. [C.N. 1]
