PROP. 1-11.

1. Divide a given straight line into two such parts that the rectangle contained by them may be the greatest possible.

2. The sum of the squares of two straight lines is never less than twice the rectangle contained by the straight lines.

3. Divide a given straight line into two parts such that the squares of the whole line and one of the parts shall be equal to twice the square of the other part. 4. Given the sum

of two straight lines and the difference of their squares, to find the lines.

5. In any triangle the difference of the squares of the sides is equal to the rectangle contained by the sum and difference of the parts into which the base is divided by a perpendicular from the vertical angle.

6. Divide a given straight line into such parts that the sum of their squares may be equal to a given square.

7. If ABCD be any rectangle, A and C being opposite angles, and O any point either within or without the rectangle—0A+ OC?

OB? + OD?. 8. Let the straight line AB be divided into any two parts in the point C. Bisect CB in D, and take a point E in AC such that EC

CD. Then shall ADP = AE? + ACCB. 9. If a point C be taken in AB, and AB be produced to D so that BD and AC are equal, show that the squares described upon AD and AC together exceed the square upon AB by twice the rectangle contained by AE and AC.

10. From the hypothenuse of a right-angled triangle portions are cut off equal to the adjacent sides. Show that the square on the middle segment is equal to twice the rectangle under the extreme segments.

11. If a straight line be divided into any number of parts, the square of the whole line is equal to the sum of the squares of the parts, together with twice the rectangles of the parts taken two and two together.

12. If ABC be an isosceles triangle, and DE be drawn parallel to the base BC, cutting in D and E either the side or sides produced, and EB be joined; prove that BE? = BC · DE + CE”.

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PROP. 12–14.

13. In any triangle show that the sum of the squares on the sides is equal to twice the square on half the base, and twice the square on the line drawn from the vertex to the middle of the base.

14. If squares are described on the sides of any triangle, find the difference between the sum of two of the squares and the third square, and show from your result what this becomes when the angle opposite the third square is a right angle.

15. Show also what the difference becomes when the vertex of the triangle is depressed until it coincide with the base.

16. The square on any straight line drawn from the vertex of an isosceles triangle, together with the rectangle contained by the segments of the base, is equal to the square upon a side of the triangle.

17. If a side of a triangle be bisected, and a perpendicular drawn from the middle point of the base to meet the side, then the square of the altitude of the triangle exceeds the square upon half the base by twice the rectangle contained by the side and the straight line between the points of section of the side.

18. In any triangle ABC, if perpendiculars be drawn from each of the angles upon the opposite sides, or opposite sides produced, meeting them respectively in D, E, F, show that

BA? + AC2 + CB? = 2 AE · AC + 2 CD •CB + 2 BF · BA; all lines being measured in the same direction round the triangle.

19. Construct a square equal to the sum of the areas of two given rectilinear figures.

20. The base of a triangle is 63 ft., and the sides 25 ft. and 52 ft. respectively. Show that the segments of the base, made by a perpendicular from the vertex, are 15 ft. and 48 ft. respectively, and that the area of the triangle is 630 sq. ft.

21. In the same triangle, show that the length of the line joining the vertex with the middle of the base is 22.9 ft.

22. A ladder, 45 ft. long, reaches to a certain height against a wall, but, when turned over without moving the foot, must be shortened 6 ft. in order to reach the same height on the opposite side. Supposing the width of the street to be 42 ft., show that the height to which the ladder reaches is 36 ft.

23. The base and altitude of a triangle are 8 in. and 9 in. respectively; show that its area is equal to a square whose side is 6 in. Prove your result by construction.

24. On the supposition that lines can be always expressed exactly in terms of some unit of length, what geometrical propositions may be deduced from the following algebraical identities ?

(1.) (a + b)2 = a + 2 ab + 12
(2.) (a + b) (a - b) + b2 = a?
(3.) (2 a + b) b + a2 = (a + b)2
(4.) (a + b)2 + 62 = 2 (a + b) b + a?
(5.) 4 (a + b) b + b2 (a + 2b)?
(6.) (a + b)2 + (a - b)2 = 2 a2 + 212
(7.) (2 a + b)2 + 62 = 2 a? + 2 (4 + 2)?



1. Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circum

ferences are equal.

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4. 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.

5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre.

6. A segment of a circle is the figure contained by a straight line and the circum

ference which it cuts off. 7. The angle of a segment is that which is contained by the straight line and the circumference.

8. 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. 9. An angle is said to insist

the circumference intercepted between the straight lines that contain the angle.

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stand upon

10. A sector of a circle is the figure contained by two straight lines drawn from the centre and the circumference between them.

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[Any portion of the circumference is called an arc, and the chord of an arc is the straight line joining its extremities.]

Proposition 1.- Problem.

To find the centre of a given circle.

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Let ABC be the given circle.
It is required to find its centre.

CONSTRUCTION.—Draw within the circle any chord AB, and bisect it in D (I. 10).

From the point D draw DC at right angles to AB (I. 11).

Produce CD to meet the circumference in E, and bisect CE in F (I. 10).

Then the point F shall be the centre of the circle ABC. Proof.—For if F be not the centre, if A

Suppose possible let G be the centre; and join GA, GD, GB.

Then, because DA is equal to DB (Const.), and Du common to the two triangles ADG, BDG;

The two sides AD, DG are equal to the two sides BD, DG, each to each;

And the base GA is equal to the base GB, being radii of the same circle;

Therefore the angle ADG is equal to the angle BDG (I. 8).

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle (I. Def. 10).

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G the centre.

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:..GDB = FDB,

Therefore the angle GDB is a right angle.
But the angle FDB is also a right angle (Const.);

Therefore the angle GDB is equal to the angle FDB (Ax. 11), the less to the greater; which is impossible.

Therefore G is not the centre of the circle ABC.

In the same manner it may be shown that no point which is not in CE can be the centre.

And since the centre is in CE, it must be in F, its point of bisection.

Therefore F is the centre of the circle ABC: which was to be found.

COROLLARY.–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.

Proposition 2.-Theorem. 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 and B any two points in the circumference.

The straight line drawn from A to B shall fall within the circle. CONSTRUCTION.- Find D the centre of the circle ABC

(III. 1), and join DA, DB.

In AB take any point E; join DE, and produce it to the circumference in F.

PROOF.—Because DA is equal to DB, the angle DAB is equal to the angle DBA (1. 5).

And because AE, a side of the triangle = Z DAB,

DAE, is produced to B, the exterior angle

DEB is greater than the interior and DBE,

opposite angle DAE (I. 16). But the angle DAE was proved to be equal to the angle DBE;

Therefore the angle DEB is also greater than DBE.

But the greater angle is subtended by the greater side

(I. 19); .:DB>DE. Therefore DB is greater than DE,




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