And because the angle GBE is equal to the angle ABM, [I.15.
and likewise to the angle D ; [Construction.
the angle ABM is equal to the angle D. [Axiom 1.
Wherefore to the given straight line AB the parallelogram LB is applied, equal to the triangle C, and having the angle ABM equal to the angle D. q.e.f.
PROPOSITION 45. PROBLEM.
To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.
Let ABCD be the given rectilineal figure, and E the given rectilineal angle: it is required to describe a par- allelogram equal to ABCD, and having an angle equal to E.
Join DB, and describe the parallelogram FH equal to
the triangle ADB, and having the angle FKH equal to the angle E ; [I. 42.
and to the straight line GH apply the parallelogram GM
equal to the triangle DBC, and having the angle GHM
equal to the angle E. [I. 44.
The figure FKML shall be the parallelogram required.
Because the angle E is equal to each of the angles FKH, GHM, [Construction.
the angle FKH is, equal to the angle GHM. [Axiom 1.
Add to each of these equals the angle KHG ;
therefore the angles FKH, KHG are equal to the angles KHG,GHM. [Axiom 2.
But FKH, KHG are together equal to two right angles;[I.29.
therefore KHG,GHM are together equal to two right angles.
