Dem.—The parallelogram AM is equal to GE [xliii.]; but GE is equal to the triangle C (const.); therefore AM is equal to the triangle C. Again, the angle ABM is equal to EBG [xv.], and EBG is equal to D (const.); therefore the angle ABM is equal to D; and AM is constructed on the given line; therefore it is the parallelogram required. PROP. XLV.—Problem. To construct a parallelogram equal to a given rectilineal figure (ABCD), and having an angle equal to a given rectilineal angle (X). Sol.—Join BD. Construct a parallelogram EG [xlii.] equal to the triangle ABD, and having the angle E equal to the given angle X; and to the right line GH apply the parallelogram HI equal to the triangle BCD, and having the angle GHK equal to X [xliv.], and so on for additional triangles if there be any. Then EI is a parallelogram fulfilling the required conditions. Dem.—Because the angles GHK, FEH are each equal to X (const.), they are equal to one another: to each add the angle GHE, and we have the sum of the angles GHK, GHE equal to the sum of the angles FEH, GHE; but since HG is parallel to EF, and EH intersects them, the sum of FEH, GHE is two right angles [xxix.]. Hence the sum of GHK, GHE is two right angles; therefore EH, HK are in the same right line [xiv.]. Again, because GH intersects the parallels FG, EK, the alternate angles FGH, GHK are equal [xxix.]: to each add the angle HGI, and we have the sum of the angles FGH, HGI equal to the sum of the angles GHK, HGI; but since GI is parallel to HK, and GH intersects them, the sum of the angles GHK, HGI is equal to two right angles [xxix.]. Hence the sum of the angles FGH, HGI is two right angles; therefore FG and GI are in the same right line [xiv.]. Again, because EG and HI are parallelograms, EF and KI are each parallel to GH; hence [xxx.] EF is parallel to KI, and the opposite sides EK and FI are parallel; therefore EI is a parallelogram; and because the parallelogram EG (const.) is equal to the triangle ABD, and HI to the triangle BCD, the whole parallelogram EI is equal to the rectilineal figure ABCD, and it has the angle E equal to the given angle X. Hence EI is a parallelogram fulfilling the required conditions. It would simplify Problems xliv., xlv., if they were stated as the constructing of rectangles, and in this special form they would be better understood by the student, since rectangles are the simplest areas to which others are referred. Exercises. 1. Construct a rectangle equal to the sum of two or any number of rectilineal figures. 2. Construct a rectangle equal to the difference of two given figures. PROP. XLVI.—Problem. On a given right line (AB) to describe a square. Sol.—Erect AD at right angles to AB [xi.], and make it equal to AB [iii.]. Through D draw DC parallel to AB [xxxi.], and through B draw BC parallel to AD; then AC is the square required.