Now, since AC is bisected in P, EK is also bisected in P; hence [I. xxxvi.] the parallelogram EO is equal to OK; therefore EO is greater than DK; but DK is equal to FD [I. xliii.]; hence EO is greater than FD. To each add BO, and we have the parallelogram BP greater than BD. Hence BP is the maximum parallelogram which can be inscribed in the given triangle. Cor. 1.—The maximum parallelogram exceeds any other parallelogram about the same angle in the triangle, by the area of the similar parallelogram whose diagonal is the line between the middle point P of the opposite side and the point D, which is the corner of the other inscribed parallelogram. Cor. 2.—The parallelograms inscribed in a triangle, and having one angle common with it, are proportional to the rectangles contained by the segments of the sides of the triangle, made by the opposite corners of the parallelograms. Cor. 3.—The parallelogram AC : GH :: AC2 : AD.DC. PROP. XXVIII.—Problem. To inscribe in a given triangle (ABC) a parallelogram equal to a given rectilineal figure (X) not greater than the maximum inscribed parallelogram, and having an angle (B) common with the triangle. Sol.—Bisect the side AC opposite to B, at P. Draw PF, PE parallel to the sides AB, BC; then [xxvii.] BP is the maximum parallelogram that can be inscribed in the triangle ABC; and if X be equal to it, the problem is solved. If not, produce EP, and draw CJ parallel to PF; then describe the parallelogram KLMN [xxv.] equal to the difference between the figure PJCF and X, and similar to PJCF, and so that the sides PJ and KL will be homologous; then cut off PI equal to KL; draw IH parallel to AB, cutting AC in D, and draw DG parallel to BC. BD is the parallelogram required. Dem.—Since the parallelograms PC, PD are about the same diagonal, they are similar [xxiv.]; but PC is similar to KPT (const.); therefore PD is similar to KN, and (const.) their homologous sides, PI and KL, are equal; hence [xx.] PD is equal to KN. Now, PD is the difference between EF and GH [xxvii. Cor. 1], and KN is (const.) the difference between PC and X; therefore the difference between PC and X is equal to the difference between EF and GH; but EF is equal to PC. Hence GH is equal to X. PROP. XXIX.—Problem. To escribe to a given triangle (ABC) a parallelogram equal to a given rectilineal figure (X), and having an angle common with an external angle (B) of the triangle. Sol.—The construction is the same as the last, except that, instead of making the parallelogram KN equal to the excess of the parallelogram PC over the rectilineal figure X, we make it equal to their sum; and then make PI equal to KL; draw IH parallel to AB, and the rest of the construction as before.