| If a point is taken outside a circle and two straight lines fall from it on the circle, and if one of them cuts the circle and the other touches it, then the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the tangent. | ||
|
Let a point D be taken outside the circle ABC, and from D let the two straight lines DCA and DB fall on the circle ABC. Let DCA cut the circle ABC, and let BD touch it.
I say that the rectangle AD by DC equals the square on DB. Then DCA is either through the center or not through the center. | ||
| First let it be through the center, and let F be the center of the circle ABC. Join FB. Therefore the angle FBD is right. | III.18 | |
| And, since AC has been bisected at F, and CD is added to it, the rectangle AD by DC plus the square on FC equals the square on FD. | II.6 | |
| But FC equals FB, therefore the rectangle AD by DC plus the square on FB equals the square on FD. | ||
| And the sum of the squares on FB and BD equals the square on FD, therefore the rectangle AD by DC plus the square on FB equals the sum of the squares on FB and BD. | I.47 | |
| Subtract the square on FB from each. Therefore the remaining rectangle AD by DC equals the square on the tangent DB. | ||
| Again, let DCA not be through the center of the circle ABC. Take the center E, and draw EF from E perpendicular to AC. Join EB, EC, and ED. | III.1 | |
| Then the angle EBD is right. | III.18 | |
| And, since a straight line EF through the center cuts a straight line AC not through the center at right angles, it also bisects it, therefore AF equals FC. | III.3 | |
| Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle AD by DC plus the square on FC equals the square on FD. | II.6 | |
| Add the square on FE to each. Therefore the rectangle AD by DC plus the sum of the squares on CF and FE equals the sum of the squares on FD and FE. | ||
| But the square on EC equals the sum of the squares on CF and FE, for the angle EFC is right, and the square on ED equals the sum of the squares on DF and FE, therefore the rectangle AD by DC plus the square on EC equals the square on ED. | I.47 | |
| And EC equals EB, therefore the rectangle AD by DC plus the square on EB equals the square on ED. | ||
| But the sum of the squares on EB and BD equals the square on ED, for the angle EBD is right, therefore the rectangle AD by DC plus the square on EB equals the sum of the squares on EB and BD. | I.47 | |
| Subtract the square on EB from each. Therefore the remaining rectangle AD by DC equals the square on DB. | ||
| Therefore if a point is taken outside a circle and two straight lines fall from it on the circle, and if one of them cuts the circle and the other touches it, then the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference equals the square on the tangent. | ||
| Q.E.D. | ||
Book III Introduction - Proposition III.35 - Proposition III.37.