HE and EG are equal (47. 1.) to the square of GH: Therefore also the rectangle BE.EF, together with the square of EG, is equal to the squares of HE and EG. Take away the square of EG, which is common to both, and the remaining rectangle BE.EF is equal to the square of EH: But BD is the rectangle con tained by BE and EF, because EF is equal to ED; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A ; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square described upon EH. PROP. A. THEOR. If one side of a triangle be bisected, the sum of the squares of the other two sides is double of the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle. Let ABC be a triangle, of which the side BC is bisected in D, and DA drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA. From A draw AE perpendicular to BC, and because BEA is a right angle, AB2=(47. 1.) BE2+AE and AC2= CE2+AE2; wherefore AB2+ AC2= BE2 +CE2+2AE2. But because the line. BC is cut equally in D, and unequally in E, BE2+ CE2 (9. 2.) 2BD2 + 2DE2; therefore AB2 + AC2=2BD2 + 2DE2.2A E2. = Now DE2+AE2=(47. 1.) AD2, and 2DE2+2AE2-2AD2; wherefore AB2+ AC2-2BD2+2AD2. B D E PROP. B. THEOR. The sum of the squares of the diameters of any parallelogram is equal to the sum of the squares of the sides of the parallelogram. Let ABCD be a parallelogram, of which the diameters are AC and BD; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA. Let AC and BD intersect one another in E: and because the vertical angles AED, CEB are equal (15. 1.), and also the alternate angles EAD, ECB (29. 1.), the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each; but the sides AD and BC, which are opposite to equal angles in these triangles, are also equal (34. 1.); therefore the other sides which are opposite to the equal angles are also equal (26. 1.), viz. AE to EC, and ED to EB. Since, therefore, BD is bisected in E, AB2+AD2=(A. 2.) 2BE2+2AE2; and for the same reason, CD + BC2 : B A D E 2BE2+2EC22BE2+2A E2, because EC-AE. C Therefore AB2+AD2 +DC2+BC2=4BE2+4AE2. But 4BE2=BD2, and 4A E2=AC2 (2. Cor. 8. 2.) because BD and AC are both bisected in E; therefore AB2+ AD2+CD2+BC2=BD2+AC2. COR. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another. SCHOLIUM. In the case of the rhombus, the sides AB, BC, being equal, the triangles BEC, DEC, have all the sides of the one equal to the corresponding sides of the other, and are therefore equal: whence it follows that the angles BEC, DEC, are equal; and, therefore, that the two diagonals of a rhombus cut each other at right angles. ELEMENTS OF GEOMETRY. BOOK III. DEFINITIONS. A. The radius of a circle is the straight line drawn from the centre to the circumference. 1. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. And that line which has but one point in common with 2. Circles are said to touch one 3. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. 4. And the straight line on which the greater pe pendicular falls, is said to be farther from the centre. B. Any portion of the circumference is called an arc. The chord or subtense of an arc is the straight line which joins its two extremities. C. A straight line is said to be inscribed in a circle, when the extremities of it are in the circumference of the circle. And any straight line which meets the circle in two points, is called a secant. 5. A segment of a circle is the figure contained by a straight line, and the arc which it cuts off. 3 6. An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. An inscribed triangle, is one which has its three angular points in the circumference. And, generally, an inscribed figure is one, of which all the angles are in the circumference. The circle is said to circumscribe such a figure. 7. And an angle is said to insist or stand upon the arc intercepted between the straight lines which contain the angle. This is usually called an angle at the centre. The angles at the circumference and centre, are both said to be subtended by the chords or arcs which their sides include. 8. The sector of a circle is the figure contained by two straight lines drawn from the centre, and the arc of the circumference between them. 9. Similar segments of a circle, are those in which the angles are equal, or which contain equal an gles. PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect (10. 1.) it in D; from the point D draw (11. 1.) DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the cirele ABC. C For, if it be not, let, if possible, G be the centre, and join GA, GD, GB: Then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are radii of the same circle: therefore the angle ADG is equal (8. 1.) to the angle GDB: But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle (7. def. 1.) Therefore the angle GDB is a right angle: But FDB is likewise a right angle: wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impos FG A D B E sible Therefore G is not the centre of the circle ABC: In the same manner, it can be shown that no other point but F is the centre: that is, F is the centre of the circle ABC. COR. From this it is manifest that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. PROP. II. THEOR. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn from A to B shall fall within the circle. C D E B F Take any point in AB as E; find D (1. 3.) the centre of the circle ABC; join AD, DB and DE, and let DE meet the circumference in F. Then, because DA is equal to DB, the angle DAB is equal (5. 1.) to the angle DBA; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16. 1.) than the angle DAE; but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: Now to the greater angle the greater side is opposite (19. 1.); DB is therefore greater than DE: but BD 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. COR. Every point, moreover, in the production of AB, is farther from the centre than the circumference. If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and if it cut it at right angles, it will bisect it. Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F; it cuts it also at right angles. Take (1. 3.) E the centre of the circle, and join EA, EB. Then because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other: but the base EA is equal to the base EB; therefore the angle AFE is equal (8. 1.) to the angle BFE. And when a straight line standing upon another makes the adjacent angles equal to one another, each of them is a right (7. Def. 1.) angle: Therefore each of the angles AFE, BFE is a right angle; wherefore the straight line CD, drawn through the centre E A F B D |