But 2AD*+2ADXAE+2AD×DF=2ADX EF (E. 1. 2.) ... AC+BD'AB'+DC'+2ADXEF: ... AC'+BD'AB2+DC'+2ADX BC; because, as hath been shewn, EF BC. And, in like manner, may the proposition be demonstrated, by the help of E. 13. 2. if one of the perpendiculars drawn from B, C, fall within AD the less of the two parallel sides. PROP. XIII. 15, THEOREM. The square of the base of an isosceles triangle is the double of the rectangle contained by either side, and by the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the extremity of the base. If the vertical angle of the isosceles A be a right angle, the proof of the proposition is manifestly deducible from E. 47. 1. But let ABC, be an isosceles A, having its ver tical A, not a right angle: First, let A be an acute, and let BD be the perpendicular drawn from B to the opposite side AC: Then, BC' 2ACX CD. For, since BD is to AC, ... AB'+2AC × CD=AC2+BC From these equals take away the equal squares (hyp.) AB1 and AC, and there remains BC'= 2AC X CD. PROP. XIV. 16. THEOREM.. If from any point, in the circumference of the greater of two given concentric circles, two straight lines be drawn to the extremities of any diameter of the less, their squares shall be, together, the double of the squares of the two semi-diameters of the two given circles. Let ACB, PDE be two circles having a com mon centre K and from any point P, in the cir cumference of the greater, let PA, PB, be drawn to the extremities A and B, of any diameter AKB, of the less circle: Then PA +PB’=2KA*+2KP2, KA being a semidiameter of the less, and KP a semidiameter of the greater circle. 2 From P draw (E. 12. 1.) PF1 to AB; and, first, let PF fall without AB. And, because PF is to AB, also, ... PB+2BKXKF= KP2+KB' (E. 13. 2. ) PA =KP'+KA*+2KAXKF; wherefore, since (E, 15. def. 1.) KB=KA, if to the two former equals, the two latter be added, and if the equal rectangles, 2BKX KF, and 2KA X KF, be taken from the equal aggregates, it is manifest that PA*+PB*=2KA*+2KP2. And, in like manner, the proposition may be demonstrated, when the perpendicular PF falls within AB, the diameter of the lesser circle. A SUPPLEMENT TO THE ELEMENTS OF EUCLID. BOOK III. PROP. I. 1. THEOREM. If two circles cut each other, the straight line joining their two points of intersection is bisected, at right angles, by the straight line joining their centres. Let the circle ABC, of which the centre is K, and the circle ADC, of which the centre is L, cut one another in the points A, C; and let K, L, and A, C be joined: AC is bisected, at right angles, in E, by KL. For, join K, A, and K, C, and L, A, and L, C: And since (E. def. 15. 1.) KA= KC and LA= LC, and that KL is common to the two KAL, KCL, .'. (E. 8. 1.) the ≤ AKL=2 CKL. Again, since AK CK, and that KE is common to the two ▲ AEK, CEK, and, as hath been shewn, the LAKE=CKE, .. (E. 4. 1.) AE=CE, and the AEK=/ CEK, so that (E. def. 10. 1.) each of these is a right 4. Wherefore, KL bisects AC at right angles. 2. COR. Hence, if a trapezium have two of its adjacent sides equal to one another, and also its two remaining sides equal to one another, its diameters bisect each other at right angles. 3. PROBLEM. PROP. II. Through a given point within a circle, which is not the centre, to draw a chord which shall be bisected in that point. Let A be a given point within the circle BCD: It is required to draw, through A, a chord of the circle BCD, which shall be bisected in the point A. Find (E. 1.3.) the centre K of the circle BCD; join A, K; draw (E. 11. 1.) through A, the chord BAC to KA: Then is BC bisected in the given point A. |