Cor.—The parallelograms AF, FC, AC are, two by two, homothetic. PROP. XXV.—Problem. To describe a rectilineal figure equal to a given one (A), and similar to another given one (BCD). Sol.—On any side BC of the figure BCD describe the rectangle BE equal to BCD [I. xlv.], and on CE describe the rectangle EF equal to A. Between BC, CF find a mean proportional GH, and on it describe the figure GHI similar to BCD [xviii.], so that BC and GH may be homologous sides. GHI is the figure required. Dem.—The three lines BC, GH, CF are in continued proportion; therefore BC : CF in the duplicate ratio of BC : GH [V. Def. x.]; and since the figures BCD, GHI are similar, BCD : GHI in the duplicate ratio of BC : GH [xx.]; also BC : CF :: rectangle BE : rectangle EF. Hence rectangle BE : EF :: figure BCD : GHI; but the rectangle BE is equal to the figure BCD; therefore the rectangle EF is equal to the figure GHI; but EF is equal to A (const.). Therefore the figure GHI is equal to A, and it is similar to BCD. Hence it is the figure required. Or thus: Describe the squares EFJK, LMNO equal to the figures BCD and A respectively [II. xiv.]; then find GH a fourth proportional to EF, LM, and BC [xii.]. On GH describe the rectilineal figure GHI similar to the figure BCD [xviii.], so that BC and GH may be homologous sides. GHI is the figure required. Dem.—Because EF : LM :: BC : GH (const.), the figure EFJK : LMNO :: BCD : GHI [xxii.]; but EFJK is equal to BCD (const.); therefore LMNO is equal to GHI; but LMNO is equal to A (const.). Therefore GHI is equal to A, and it is similar to BCD. PROP. XXVI.—Theorem. If two similar and similarly situated parallelograms (AEFG, ABCD) have a common angle, they are about the same diagonal. Dem.—Draw the diagonals (see fig., Prop. xxiv.) AF, AC. Then because the parallelograms AEFG, ABCD are similar figures, they can be divided into the same number of similar triangles [xx.]. Hence the triangle FAG is similar to CAD, and therefore the angle FAG is equal to the angle CAD. Hence the line AC must pass through the point F, and therefore the parallelograms are about the same diagonal. Observation.—Proposition xxvi., being the converse of xxiv., has evidently been misplaced. The following would be a simpler enunciation:—“If two homothetic parallelograms have a common angle, they are about the same diagonal.” PROP. XXVII—Problem. To inscribe in a given triangle (ABC) the maximum parallelogram having a common angle (B) with the triangle. Sol.—Bisect the side AC opposite to the angle B, at P : through P draw PE, PF parallel to the other sides of the triangle. BP is the parallelogram required. Dem.—Take any other point D in AC : draw DG, DH parallel to the sides, and CK parallel to AB; produce EP, GD to meet CK in K and J, and produce HD to meet PK in I.