Kitap 1
44
But BF is equal to the triangle C; therefore LB is also equal to C. [C.N. 1]
44
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.
44
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.
44
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.
45
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]
45
Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM.
45
But the angles FKH, KHG are equal to two right angles; [I. 29] therefore the angles KHG, GHM are also equal to two right angles.
