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11. The three lines drawn from the angles of a triangle to the middle of the opposite sides all pass through the same point, and three times the sum of the squares on the lines intercepted between this point and the angles is equal to the sum of the squares on the three sides of the triangle.

12. If the three sides of a triangle be 4, 8, and 10, find by Proposition c, Scholium, whether the triangle is acute or obtuse-angled.

13. If perpendiculars be drawn from the three angles of a triangle to the opposite sides, prove that the sum of the squares on the alternate segments of the sides are equal to one another.

14. If from one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side, twice the rectangle contained by that side and the segment of it, between the base and perpendicular, is equal to the square on the base.

15. The square on the base of an isosceles triangle, whose vertical angle is a right angle, is equal to four times the area of the triangle.

16. If in the figure to (I. 47), the angular points of the squares be joined, thus forming a six-sided figure, prove that the sum of the squares on the six sides is equal to eight times the square on the hypotenuse of the original triangle.

17. The sum of the squares on the sides of a quadrilateral is equal to the sum of the squares on its diagonals, and four times the square on the line joining their middle points.

18. The sum of the squares on two opposite sides of a quadri. lateral, together with four times the square on the line joining their middle points, is equal to the sum of the squares on the other two sides and on the diagonals.

19. The squares on the sum, and on the difference of two lines, are together double the squares on these lines.

20. If a line be cut in medial section, the line composed of it and its greater segment is similarly divided.

21. The sum of the squares on the diagonals of a quadrilateral, is equal to twice the sum of the squares on the lines joining the middle points of the opposite sides.

22. If the vertical angle of a triangle be four-thirds of a right angle, the square on the base is equal to the sum of the squares on the sides, together with the rectangle contained by the sides ; and if the vertical angle be two-thirds of a right angle, the square on the base is equal to the sum of the squares on the two sides diminished by the rectangle contained by the sides.

THIRD BOOK.

DEFINITIONS.

1. An arc of a circle is any portion of the circumference, as abc.

2. The chord of an arc is a straight line joining its extremities, as ac.

3. The arc of a semicircle is called a semicircumference; and a radius is called a semidiameter. 14. A segment of a circle is a figure contained by an arc and its chord, as S.

5. An angle in a segment is an angle contained by two lines, drawn from any point in its arc, to the extremities of its chord ; x is the angle in the segment ABC. ... 6. An angle is said to insist or stand upon the arc which subtends it. 157. A sector of a circle is a figure contained by two radii and the intercepted arc, as S. When the radii are perpendicular, the sector is called a quadrant.

18. Similar segments of a circle are those that contain equal angles, as S, s'.

9. Similar arcs of circles are those that subtend equal angles at the centre.

10. Similar sectors are those that are bounded by similar arcs.

11. Equal circles are those that have equal radii; and concentric circles are those that have a common centre, c.

12. 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, as AB.

13. Circles are said to touch one another when they meet, but do not cut one another. Such circles may be called tangent circles.

14. The point in which a tangent and a circle, or two tangent circles, meet, is called the point of contact ; C, C, c, are points of contact.

15. Chords are said to be equally distant from the centre of a circle, when the perpendiculars upon them, from the centre, are equal; and the chord on which the greater perpendicular falls, is said to be farther from the centre.

16. A secant is a straight line drawn from a point, without a circle, and meeting it in two points. A secant may be considered a chord produced, the point without the circle from which it is drawn being a point of external section.

PROPOSITION I. PROBLEM.
To find the centre of a given circle.
Given the circle ABC; it is required to find its centre.

(Const.) Draw within it any chord AB, and bisect it in D (I. 10); from the point D draw DC at right angles to AB (I. 11), and produce it to E, and bisect CE in F. The point F is the centre of the circle ABC.

(Dem.) For, if it be not, let, if possible, G, a point not in CE, 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 and DG are equal to the two BD and DG, each to each; and the radius GA is equal to the radius GB; therefore the angle ADG is equal to the angle GDB (I. 8);

therefore (I. Def. 11) 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 impossible; therefore G is not the centre of the circle ABC. In the same manner it can be shewn, that no point which is not in CE can be the centre; that is, the centre is in CE;

and being in CE, it must be in its point of bisection; therefore 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.

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PROPOSITION II. THEOREM. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Given a circle ABC, and A, B any two points in the circumference; to prove that the straight line drawn from A to B shall fall within the circle.

(Const.) Take any point in AB as E; find D the centre of the circle ABC; join AD, DB, and DE, and let DE meet the circumference in F. (Dem.) Then, because DA is equal to DB, the angle DAB is equal to the angle DBA (I. 5); and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater than the angle DAE (I. 16); but DAE is equal to the angle DBE, therefore the angle DEB is greater than the angle DBE;

DB is therefore greater than DE (I. 19). 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.

Schol. When the two points are at the extremities of a diameter, or diametrically opposite, the truth of the proposition is evident. It is obvious, from the nature of the circle, when they are at a less distance. When the points are exceedingly near, however, it is not then so evident that the chord lies wholly within the circle, and no part of it on the arc; and therefore, in this case, it is necessary to demonstrate the proposition. Some parts of the figure, however, in this extreme case, would be so small as to be indistinct, and therefore it is necessary to assume the points at a sufficient distance; and as the reasoning in this case applies to every case, the proposition will therefore be established also in the extreme case. A similar observation applies to several propositions, which appear to be axiomatic, except in the extreme case, as I. 20.

PROPOSITION III. THEOREM. If a straight line drawn through the centre of a circle bisect a chord in it which does not pass through the centre, it shall cut it at right angles; and if it cuts it at right angles, it shall bisect it.

Given a circle ABC, and CD, a straight line drawn through the centre, bisecting any chord AB, which does not pass through the centre, in the point F; to prove that it cuts it also at right angles.

(Const.) Take E the centre of the circle (III. 1), and join EÀ, EB. (Dem.) Then, because AF is equal to FB, and FE common to the two triangles AFE and BFE,

there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE (I. 8); therefore (I. Def. 11) each of the angles AFE and BFE is a right angle; wherefore the straight line CD, drawn through the centre bisecting another AB that does not pass through the centre, cuts the same at right angles. Second,

Given that CD cuts AB at right angles; to prove that CD also bisects AB; that is, that AF is equal to FB.

The same construction being made, (Dem.) because EA and EB from the centre are equal to one another, the angle EAF is equal to the angle EBF (I. 5); and the right angle AFE is equal to the right angle BFÉ; therefore, in the two triangles EAF and EBF, there are two angles in one equal to two angles in the other, and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal (I. 26); AF is therefore equal to FB. COR.-If a line bisect a chord of a circle, and be perpendicular

to it, it will pass through the centre. For a line from the centre bisecting the chord is perpendicular to it, and therefore coincides with the former perpendicular;

therefore it must also pass through the centre. Schol.-It appears, therefore, that of the three conditions-of passing through the centre of a circle of being perpendicular to a chord, and of bisecting a chord; if a line fulfil two, it will also fulfil the third.

PROPOSITION IV. THEOREM. If in a circle two chords cut one another which do not both pass through the centre, they do not bisect each other.

Given a circle ABCD, and AC, BD, two chords in it, which cut one another in the point E, and do not both pass through the centre; to prove that AC and BD do not bisect one another.

(Dem.) For, if it be possible, let AE be equal to EC, and BE to ED. If one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre. But if neither of them pass through the centre, take F the centre of the circle, and join EF (III. 1);

and because FE, a straight line through the centre, bisects

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