Dem.—Through E draw EM parallel to BC; make EN = EM; produce AB until EP = AB. Through the points N, P draw NT, RP each parallel to AC, and through P draw PQ parallel to BC. It is easy to see from the figure that the parallelogram QR is equal to the parallelogram MF, and is therefore given. Through P draw ST perpendicular to EP. Now, since AE = DB, BP is equal to DB; therefore PS = CD. Again, since OP = AD, PT is equal to CD; therefore PS = PT. Hence QR is the maximum parallelogram in the triangle SV T. Again, if any other line A′B be drawn through E, and produced to P′, so that EP′ = AP′, the point P′ must fall outside ST, because the parallelogram Q′R′, corresponding to QR, will be equal to MF, and therefore equal to QR. Hence the line EP′ is greater than EP, or A′B′ is greater than AB. Hence AB is a minimum. ________________ NOTE F. on the trisection of an angle. The following mechanical method of trisecting an angle occurred to me several years ago. Apart from the interest belonging to the Problem, it is valuable to the student as a geometrical exercise:— To trisect a given angle ACB. Sol.—Erect CD perpendicular to CA; bisect the angle BCD by CG, and make the angle ECI equal half a right angle; it is evident that CI will fall between CB and CA. Then, if we use a jointed ruler—that is two equal rulers connected by a pivot—and make CB equal to the length of one of these rulers, and, with C as centre and CB as radius, describe the circle BAM, cutting CI in I: at I draw the tangent IG, cutting CG in G. Then, since ICG is half a right angle, and CIG is right, IGC is half a right angle; therefore IC is equal to IG; but IC equal CB; therefore IG = CB—equal length of one of the two equal rulers. Hence, if the rulers be opened out at right angles, and placed so that the pivot will be at I, and one extremity at C, the other extremity at G; it is evident that the point B will be between the two rulers; then, while the extremity at C remains fixed, let the other be made to traverse the line GF, until the edge of the second ruler passes through B: it is plain that the pivot moves along the circumference of the circle. Let CH, HF, be the positions of the rulers when this happens; draw the line CH; the angle ACH is one-third of ACB. Dem.—Produce BC to M. Join HM. Erect BO at right angles to BM. Then, because CH = HF, the angle HCF = HFC, and the angle DCE = ECB (const.). Hence the angle HCD = HBC [I. xxxii.], and the right angles ACD, CBO are equal; therefore the angle ACH is equal to HBO; that is [III. xxxii.], equal to HMB, or to half the angle HCB. Hence ACH is one-third of ACB. ________________ NOTE G. on the quadrature of the circle.