Problems.

OF THE RULER.

A ruler of a convenient size, is about twenty inches in ength, two inches wide, and one fifth of an inch in thickness. t 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 the same edge falls on B. Then, with a pen, or pencil, draw the line 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 B as a centre, and an equal radius BI describe the arc IHE. the points I and E by the line IE:

Join

the point D, where it intersects AB, will be the middle point of the line AB.

A

B

H

Problems.

For, draw the radii AI, AE BI, and BE. Then, since these radii are equal, the triangles AIE and BIE have all the sides of the one equal to the corresponding sides of the other; hence, their corresponding 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 th

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 arc AEB at E.

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

and CD common: hence, the two triangles have their corresponding angles equal (Bk I. Th. viii), and consequently, ACD is equal to BCD. But since ACD is equal to BCD, it follows 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

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

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

SECOND METHOD.

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

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

D

B C

A

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

Problems.

PROBLEM VII.

From a given port without a straight line to let fall a perpen

the same radius, describe two arcs intersecting each other at E. Draw AFE, and it will be the perpendicular required.

For, draw the equal radii AB, AD, BE and DE. Then, the two triangles EAB and EAD will have the sides of the one equal to the sides of the other, each to each; hence, their corresponding angles will be equal (Bk. I. Th. viii), viz. the angle BAE to the angle DAE. Hence, the two triangles BAF and DAF will have two sides and the included angle of the one, equal to two sides and the included angle of the other, and therefore, the angle AFB will be equal to the angle AFD (Bk. I. Th. iv): hence, AFE will be perpendicular to BD.

SECOND METHOD.

When the given point A is nearly opposite the extremity of the line.

Draw AC, to any point C of the line BD. Bisect AC at P. Then, with P as a centre and PC as a radius, describe the semicircle CDA; draw AD, and it will be perpendicular

B C

P

to CD since CDA is an angle in a semicircle (Bk. II. Th. x).

Prob em s.

PROBLEM VIII.

At a given point in a given line, to make an angle equal to a

given angle.

A A

angle.

From the vertex K, as a centre,

with any radius, describe the arc IL, terminating in the two sides of the angle: and draw the chord IL.

From the point A, as a centre, with a distance AE, equal to KI, describe the arc DE; then with E, as a centre, and a radius equal to the chord IL, describe an arc cutting DE at D; draw AD, and the angle EAD will be equal to the angle K.

For, draw the chord DE. Then the two triangles IKL and EAD, having the three sides of the one equal to the three sides of the other, each to each, the angle EAD will be equal to the angle K (Bk. I. Th. viii).

PROBLEM IX.

Through a gwen point to draw a line that shall be parallel to a given line.

Let A be the given point and

BC the given line.

With A as a centre, and any ra

dius greater than the shortest dis

B

tance from A to BC, describe the indefinite arc DE.

D

From

the point E, as a centre, with the same radius, describe the arc AF: then, make ED equa to AF and draw AD, and it will be the required parallel.