To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.

Let *ABCD* be the given rectilinear figure and *E* the given rectilinear angle.

It is required to construct a parallelogram equal to the rectilinear figure *ABCD* in the given angle *E.*

Join *DB.* Construct the parallelogram *FH* equal to the triangle *ABD* in the angle *HKF* which equals *E.* Apply the parallelogram *GM* equal to the triangle *DBC* to the straight line *GH* in the angle *GHM* which equals *E.*

Since the angle *E* equals each of the angles *HKF* and *GHM,* therefore the angle *HKF* also equals the angle *GHM.*

Add the angle *KHG* to each. Therefore the sum of the angles *FKH* and *KHG* equals the sum of the angles *KHG* and *GHM.*

But the sum of the angles *FKH* and *KHG* equals two right angles, therefore the sum of the angles *KHG* and *GHM* also equals two right angles.

Thus, with a straight line *GH,* and at the point *H* on it, two straight lines *KH* and *HM* not lying on the same side make the adjacent angles together equal to two right angles, therefore *KH* is in a straight line with *HM.*

Since the straight line *HG* falls upon the parallels *KM* and *FG,* therefore the alternate angles *MHG* and *HGF* equal one another.

Add the angle *HGL* to each. Then the sum of the angles *MHG* and *HGL* equals the sum of the angles *HGF* and *HGL.*

But the sum of the angles *MHG* and *HGL* equals two right angles, therefore the sum of the angles *HGF* and *HGL* also equals two right angles. Therefore *FG* is in a straight line with *GL.*

Since *FK* is equal and parallel to *HG,* and *HG* equal and parallel to *ML* also, therefore *KF* is also equal and parallel to *ML,* and the straight lines *KM* and *FL* join them at their ends. Therefore *KM* and *FL* are also equal and parallel. Therefore *KFLM* is a parallelogram.

Since the triangle *ABD* equals the parallelogram *FH,* and *DBC* equals *GM,* therefore the whole rectilinear figure *ABCD* equals the whole parallelogram *KFLM.*

Therefore the parallelogram *KFLM* has been constructed equal to the given rectilinear figure *ABCD* in the angle *FKM* which equals the given angle *E.*

Q.E.F.

With this construction any rectilinear area can be applied to a line in an angle, that is, it can be transformed into a parallelogram with whatever angle you want and with one side whatever you want. That is a satisfactory solution to the question “what’s the area of this figure?”

But the question “what’s the area of a circle?” is not answered in the *Elements.* See the note on squaring the circle after proposition II.14 for more discussion of this question.