To a given straight line in a given rectilinear angle, to apply a parallelogram equal to a given triangle.

Let *AB* be the given straight line, *D* the given rectilinear angle, and *C* the given triangle.

It is required to apply a parallelogram equal to the given triangle *C* to the given straight line *AB* in an angle equal to *D.*

Construct the parallelogram *BEFG* equal to the triangle *C* in the angle *EBG* which equals *D,* and let it be placed so that *BE* is in a straight line with *AB.*

Draw *FG* through to *H,* and draw *AH* through *A* parallel to either *BG* or *EF.* Join *HB.*

Since the straight line *HF* falls upon the parallels *AH* and *EF,* therefore the sum of the angles *AHF* and *HFE* equals two right angles. Therefore the sum of the angles *BHG* and *GFE* is less than two right angles. And straight lines produced indefinitely from angles less than two right angles meet, therefore *HB* and *FE,* when produced, will meet.

Let them be produced and meet at *K.* Draw *KL* through the point *K* parallel to either *EA* or *FH.* Produce *HA* and *GB* to the points *L* and *M.*

Then *HLKF* is a parallelogram, *HK* is its diameter, and *AG* and *ME* are parallelograms, and *LB* and *BF* are the so-called complements about *HK.* Therefore *LB* equals *BF.*

But *BF* equals the triangle *C,* therefore *LB* also equals *C.*

Since the angle *GBE* equals the angle *ABM,* while the angle *GBE* equals *D,* therefore the angle *ABM* also equals the angle *D.*

Therefore the parallelogram *LB* equal to the given triangle *C* has been applied to the given straight line *AB,* in the angle *ABM* which equals *D.*

Q.E.F.

To “apply an area to a line in an angle” means just what this construction accomplishes, namely, to construct a parallelogram equal to that area with one side as the given line and one angle equal to the given angle.

In practice the angle is often a right angle. The given line may be thought of as a unit line. Then the length of the resulting rectangle represents the the area.