the squares of AB and AC are equal to twice the squares of BD and AD. Therefore the squares, &c. * PROP. B. THEOR. The squares of the diagonals of a parallelogram are together equal to the squares of its sides. Let ABCD be a parallelogram, of which the diagonals are AC and BD: the sum of the squares of AC and BD is equal to the sum of the squares of the sides AB, BC, CD, DA. Let AC and BD intersect each other in E. Then, in the triangles AED, CEB, the angles AED, ADE are respectively equal (I. 15. and 29.) to CEB, CBE, and (I. 34.) the side AD is equal to CB; therefore the sides AE, ED are respectively equal (I. 26.) to CE, EB. Since, therefore, BD is bisected in E, the squares of AB, AD are together equal (II. A.) to twice the squares of AE, EB. But BC and CD are respectively equal (I. 34.) to AD and AB; and therefore the squares of AB, BC, CD, and DA are together equal to four times the squares AE and EB, that is, (II. 4. cor. 2.) to the squares of AC and BD, since AC is double of AE, and BD of BE. Therefore, &c. + Cor. From this demonstration, it is manifest, that the diagonals of every parallelogram bisect one another. B Е. D of PROP. C. THEOR. The sum of the squares of the sides of a trapezium is equal to the sum of the squares of the diagonals, together with four times the square of the straight line joining the points of bisection of the diagonals. Let ABCD be a trapezium, having its diagonals AC, BD bisected in E and F, and let EF be joined : the squares of AB, BC, CD, DA are together equal to the squares of AC, BD, together with four times the square of EF. Join AF, FC. The squares of AB, AD are together equal (II. A) to twice the sum of the squares of DF and AF; and the squares of BC, B F A D • Hence, if the sides of a triangle be given in numbers, the line AD can be computed. Thus if AB= I1, BC= 14, and AC=7, we have AB? + AC2 = 121 + 49= 170, and 2BD2 = 98. Then 170-98=72, the ball of which is 36; and 6, the square root of this, is AD. † Hence, if we have the sides and one of the diagonals of a parallelogram in numbers, we can compute the remaining diagonal. Thus, if AB, DC be each = 9, AD, BC each :7, and AC= 8, we have AB? + BC? + CD? + DA? = 81 + 49 +81 + 49= 260, and AC2 = 64. Taking the latter from the former, and extracting the square root, we find BD= 14. B CD are equal to twice the sum of the squares of DF and CF. Add these equals together, and the sum of the squares of AB, BC, CD, DA is equal to four times the square of DF, together with twice the sum of the squares of AF and CF. But twice the squares of AF and CF are equal (II. A.) to four times the squares of AE and EF; and (II. 4. cor. 2.) four A times the square of DF is equal to the square of BD, and four times the square of AE to the square of AC. Hence the squares of AB, BC, CD, DA are equal to the squares of AC and BD, together with four times the square of EF. Therefore the sum, &c. с Proposition B. may be regarded as a particular case of this proposition, the line EF vanishing in the parallelogram, since the diagonals bisect each other ; and thus the proof of the former may be regarded as contained in this. In all such cases, however, Euclid gives separate proofs ; and it seems better to do so in an elementary work. BOOK III.* DEFINITIONS. + 1. A STRAIGHT line is said to touch a circle, or to be a tangent to it, when it meets the circle, and being produced does not cut it. 2. Circles are said to touch one another, which meet, but do not cut, one another. 3. In a circle, chords are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal : 4. And the chord on which the greater perpendicular falls, is said to be farther from the centre. 5. An angle in a segment of a circle is an angle contained by two straight lines drawn from any point in the arc of the segment to the extremities of its chord : 6. And an angle is said to stand upon the arc intercepted between the straight lines that contain the angle. 7. A sector of a circle is a figure contained by any arc of the circle, and two radii drawn through its ex tremities. I * The third book of the Elements treats of the properties of the circle, and of the lines and angles dependent on it. + The first definition in Simson's edition is as follows: “ Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal.” This is here omitted, because, as Simson justly remarks, it “is not a definition but a theorem, the truth of which is evident ; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal.” The definition of a segment is also omitted, as it is given in the first book. † As any chord divides a circle into two segments, so any two radii divide it into two sectors. A quadrant is a sector whose radii are perpendicular to each other. It is easy to show by superposition, that a quadrant is half of a semicircle, and therefore a fourth part of the entire circle. 8. Similar segments of circles are those which contain equal angles. 9. Concentric circles are those which have the same centre. Draw any B PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. chord AB, and (I. 10.) bisect it in D; from D draw (I. 11.) EDC at right angles to AB, and bisect it in F: F is the centre. * For, if it be not, let, if possible, G, a point not in CE, be the centre, and join GA, GD, GB. Then, in the triangles ADG, BDG, because DA is equal (const.) to DB, DG common, and GA equal (I. def. 30.) to GB, because they are drawn from the centre G to the circumference: therefore (I. 8.) the angle ADG is equal to GDB. But (I. def. 8.) when one straight line standing on another, makes the adjacent angles equal, each of the angles is a right angle: therefore GDB is a right angle: but (const.) FDB is likewise a right angle ; wherefore FDB is equal to GDB, the greater to the less, which is impossible : therefore G is not the centre of the circle. In the same manner it can be shown, that no other point, which is not in CE, is the centre: the centre is therefore in CE: and since FC is equal to FE, F is the centre : which was to be found. + Cor. From this it is manifest, that a straight line which bisects a chord of a circle at right angles, passes through the centre. Schol. The practical construction is very simply effected in the following manner: Take any three points A, B, C, in the circumference, and from A and B as centres, with any radius greater than half their distance, describe arcs intersecting each other in D and E; also from B and C as centres, with any radius greater than half their distance, describe arcs intersecting in F and G: through E and D, and through G and F, draw straight lines intersecting each other in H: H is the centre. The proof is evident from the preceding corollary; for EH and GH are evidently perpendicular to chords D EX B In Simson's edition, the construction commences thus, “ Draw within the circle any straight line AB.” This mode of expression is improper, as it is not to be assumed till the second proposition is proved, that AB falls within the circle. In like manner, we are directed to produce DC to E, which also implies what is proved in the second proposition. + The demonstration of this proposition which is given in the editions of Simson, Playfair, and others, is imperfect; as, according to it, any point what. ever in CE, or its continuations, would appear to be the centre. joining AB and BC, and bisect them, and therefore the centre must be in both EH and GH; it must therefore be their point of intersection, which is the only point common to both. PROP. II. THEOR. The straight line which joins any two points in the circumference of a circle, falls within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the chord drawn from A to B falls within the circle. Take any point in A B as E; find (III. 1.) D the centre of the circle ABC ; join AD, DB, DE, and let DE meet the circumference in F. Then, because DA is equal to DB, the angles DAB, DBA are (I. 5.) equal; and because AE, a side of the triangle DAÈ, is produced to B, the angle DEB is greater (I. 16.) than DAE; but DAE is equal to DBE; therefore the angle DEB is greater than DBE. Now, (I. 19.) to the greater angle the greater side is opposite; DB is therefore greater than DE: but DB is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B; therefore AB is within the circle. Wherefore the straight line, &c. B PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect a chord, which does not pass through the centre, it cuts it at right angles; and (2.) if it cut it at right angles, it bisects it. Let ABC be a circle ; and let ED, a straight line drawn through the centre E, bisect any chord AB, which does not pass through the centre, in the point D; it cuts it also at right angles. Join EA, EB. Then in the triangles ADE, BDE, AD is equal to DB, DE common, and (I. def. 30.) the base EA equal to the base EB; therefore (I, 8.) the angles ADE, BDE are equal ; and consequently (I. def. 8.) each of them is a right angle ; wherefore ED cuts AB at right angles. Next, let ED cut AB at right angles; ED also bisects it. The same construction being made, because the radii EA, EB are (I. def. 30.) equal, the angle EAD is equal (I. 5.) to EBD; and the right angles ADE, BDE are equal; therefore, in the two triangles EAD, EBD, there are two angles in one equal to two с B D |