Observation.—Proposition vii. affords a good illustration of the following important definition (see Sequel to Euclid, p. 13):—If a geometrical magnitude varies its position continuously according to any law, and if it retains the same value throughout, it is said to be a constant, such as the radius of a circle revolving round the centre; but if it goes on increasing for some time, and then begins to decrease, it is said to be a maximum at the end of the increase. Thus, in the foregoing figure, PA, supposed to revolve round P and meet the circle, is a maximum. Again, if it decreases for some time, and then begins to increase, it is a minimum at the commencement of the increase. Thus PE, supposed as before to revolve round P and meet the circle, is a minimum. Proposition viii. will give other illustrations. PROP. VIII.—Theorem. If from any point (P) outside a circle, lines (PA, PB, PC, &c.) be drawn to the concave circumference, then—1. The maximum is that which passes through the centre. 2. Of the others, that which is nearer to the one through the centre is greater than the one more remote. Again, if lines be drawn to the convex circumference—3. The minimum is that whose production passes through the centre. 4. Of the others, that which is nearer to the minimum is less than one more remote. 5. From the given point (P) there can be drawn two equal lines to the concave or the convex circumference, both of which make equal angles with the line passing through the centre. 6. More than two equal lines cannot be drawn from the given point (P) to either circumference. Dem.—1. Let O be the centre. Join OB. Now since O is the centre, OA is equal to OB: to each add OP, and we have AP equal to the sum of OB, OP; but the sum of OB, OP is greater than BP [I. xx.]. Therefore AP is greater than BP. 2. Join OC, OD. The two triangles BOP, COP have the side OB equal to OC, and OP common, and the angle BOP greater than COP; therefore the base BP is greater than CP [I. xxiv.]. In like manner CP is greater than DP, &c. 3. Join OF. Now in the triangle OFP the sum of the sides OF, FP is greater than OP [I. xx.]; but OF is equal to OE [I. Def. xxx.]. Reject them, and FP will remain greater than EP. 4. Join OG, OH. The two triangles GOP, FOP have two sides GO, OP in one respectively equal to two sides FO, OP in the other; but the angle GOP is greater than FOP; therefore [I. xxiv.] the base GP is greater than FP. In like manner HP is greater than GP. 5. Make the angle POI equal POF [I. xxiii.]. Join IP. Now the triangles IOP, FOP have two sides IO, OP in one respectively equal to two sides FO, OP in the other, and the angle IOP equal to FOP (const.); therefore [I. iv.] IP is equal to FP. 6. A third line cannot be drawn from P equal to either of the lines IP, FP. For if possible let PK be equal to PF; then PK is equal to PI—that is, one which is nearer to the minimum equal to one more remote—which is impossible.