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.
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.