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Of the Circle.

follows that the sum of the two angles A and B will be equal to one hundred and twenty degrees. But the triangle CAB is isosceles: hence, the angles at the base are equal (Bk. I. Ih. vi): hence, each angle is equal to sixty degrees, and consequently, the side AB is equal to AC or CB (Bk. I. Th vi).

PROBLEMS

RELATING TO THE FIRST AND SECOND BOOKS.

THE Problems of Geometry explain the methods of con structing or describing the geometrical figures.

For these constructions, a straight ruler and the common compasses or dividers, are all the instruments that are absolutely necessary.

DIVIDERS OR COMPASSES.

a

The dividers consist of the two legs ba, be, which turn easily about a common joint at b. The legs of the dividers

Problems.

are extended or brought together by placing the forefinger on the joint at b, and pressing the thumb and fingers against the logs

PROBLEM 1.

On any line, as CD, to lay off a distance equal to AB.

Take up the dividers with the thumb and second finger, and place the forefinger on the joint at b.

Then, set one foot of the dividers at A, and extend the legs with the thumb and fingers, until the other foot reaches B.

B

C

E

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Then, raise the dividers, place one foot at C, and mark with the other the distance CE: and this distance will evidently be equal to AB.

PROBLEM 11.

To describe from a given centre the circumference of a circle having a given radius.

Let C be the given centre, and CB the given radius.

Place one foot of the dividers at C and extend the other leg until it reaches to B. Then, turn the dividers around the leg at C, and the other leg will describe the required circumference

Problems.

OF THE RULER.

A ruler of a convenient size, is about twenty inches in length, two inches wide, and one fifth of an inch in thickness. It should be made of a hard material, and perfectly straight and smooth.

PROBLEM III.

To draw a straight line through two given points A and B. Place one edge of the ruler on

A and slide the ruler around until he same edge falls on B. Then, with a pen, or pencil, draw the ine AB.

A

B

PROBLEM IV.

To bisect a given line: that is, to divide it into two equal parts.

Let AB be the given line to be divided. With A as a centre, and radius greater than half of AB, describe an arc IFE. Then, with Bas a centre, and an equal radius

BI describe the arc IHE. Join the points I and E by the line IE. the point D, where it intersects AB, will be the middle point of the line AB.

B

H

Problems.

For, draw the radii AI, AE BI, and BE. Then, since these radii are equal, the triangles AIE nd BIE have all the sides of the one equal to the corresponding sides of the other; hence, their corres ponding angles are equal (Bk I.

A

B

Th. viii); that is, the angle AIE is equal to the angle BIE Therefore, the two triangles AID and BID, have the side AI=IB, the angle AID=BID, and ID common: hence they are equal (Bk. I. Th. iv), and AD is equal to DB.

PROBLEM V.

To bisect a given angle or a given arc.

Let ACB be the given angle, and AEB the given arc.

From the points A and B, as centres, describe with the same radius two arcs cutting each other in D. Through D and the centre • C, draw CED, and it will divide

B

the angle ACB into two equal parts, and also bisect the are AEB at E.

For, draw the radii AD and B.D. Then, in the two triangles ACD, CBD, we have

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and CD common: hence, the two triangles have their corres ponding angles equal (Bk I. Th. viii), and consequently, ACD is equal to BCD. But since ACD is equal to BCD, it fol lows that the arc AE, which measures the former, is equal to the arc BE. which measures the latter

Problems.

PROBLEM VI.

At a given point in a straight line to erect a perpendicular to the line.

Let A be the given point, and BC

the given line.

From A lay off any two distances, AB and AC, equal to each other Then, from the points B and C, as centres, with a radius greater than

B

A

AB, describe two arcs intersecting each other at D; draw DA, and it will be the perpendicular required.

For, draw the equal radii BD, DC. Then, the two trian gles, BDA, and CDA, will have

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and AD common: hence, the angle DAB is equal to the angle DAC (Bk. I. Th. viii), and consequently, DA is perpendicular to BC. (Bk. I Def. 21).

SECOND METHOD.

When the point A is near the extremity of the linc.

Assume any centre, as P, out of

the given line. Then with P as a
centre, and radius from P to A, de-
scribe the circumference of a circle
Through C, where the circumference
cuts BA, draw CPD. Then, through
D, where CP produced meets the BC
circumference, draw DA: then will

DA be perpendicular to B4, since CAD is an angle in a semicircle (Bk. II. Th. x).

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