Dem.—If possible let the circles be concentric, and let O be the centre of each. Join OA, and draw any other line OD, cutting the circles in the points B, D respectively. Then because O is the centre of each circle (hyp.), OB and OD are each equal to OA; therefore OB is equal to OD, which is impossible. Hence the circles cannot have the same centre. PROP. VII.—Theorem. If from any point (P) within a circle, which is not the centre, lines (PA, PB, PC, &c.), one of which passes through the centre, be drawn to the circumference, then—1. The greatest is the line (PA) which passes through the centre. 2. The production (PE) of this in the opposite direction is the least. 3. Of the others, that which is nearest to the line through the centre is greater than every one more remote. 4. Any two lines making equal angles with the diameter on opposite sides are equal. 5. More than two equal right lines cannot be drawn from the given point (P) to the 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 PB [I. xx.]. Therefore PA is greater than PB. 2. Join OD. Then [I. xx.] the sum of OP, PD is greater than OD; but OD is equal to OE [I. Def. xxx.]. Therefore the sum of OP, PD is greater than OE. Reject OP, which is common, and we have PD greater than PE. 3. Join OC; then two triangles POB, POC have the side OB equal to OC [I. Def. xxx.], and OP common; but the angle POB is greater than POC; therefore [I. xxiv.] the base PB is greater than PC. In like manner PC is greater than PD. 4. Make at the centre O the angle POF equal to POD. Join PF. Then the triangles POD, POF have the two sides OP, OD in one respectively equal to the sides OP, OF in the other, and the angle POD equal to the angle POF; hence PD is equal to PF [I. iv.], and the angle OPD equal to the angle OPF. Therefore PD and PF make equal angles with the diameter. 5. A third line cannot be drawn from P equal to either of the equal lines PD, PF. If possible let PG be equal to PD, then PG is equal to PF—that is, the line which is nearest to the one through the centre is equal to the one which is more remote, which is impossible. Hence three equal lines cannot be drawn from P to the circumference. Cor. 1.—If two equal lines PD, PF be drawn from a point P to the circumference of a circle, the diameter through P bisects the angle DPF formed by these lines. Cor. 2.—If P be the common centre of circles whose radii are PA, PB, PC, &c., then—1. The circle whose radius is the maximum line (PA) lies outside the circle ADE, and touches it in A [Def. iv.]. 2. The circle whose radius is the minimum line (PE) lies inside the circle ADE, and touches it in E. 3. A circle having any of the remaining lines (PD) as radius cuts ADE in two points (D, F).