PROBLEMS.

PROBLEM I.

To bisect a line AB; that is, to divide it into two equal parts.

From the two centres A and B, with any equal radii, describe arcs of circles, intersecting each other in c and D; and draw

the line CD, which will bisect the given line A

AB in the point E.

For, draw the radii AC, BC, AD, BD.

Then, because all these four radii are equal,

and the side CD common, the two triangles

B

ACD, BCD, are mutually equilateral consequently they are also mutually equiangular (th. 5), and have the angle ACE equal to the angle BCE.

Hence, the two triangles ACE, BCE, having the two sides AC, CE, equal to the two sides BC, CE, and their contained angles equal, are identical (th. 1), and therefore have the side AE equal to EB. 2. E. D.

A

From the centre A, with any radius, describe an arc, cutting off the equal lines

AD, AE; and from the two centres D, E, D

E

with the same radius, describe arcs inter

secting in F; then draw AF, which will B bisect the angle a as required.

A very ingenious instrument for trisecting an angle, is described in

the Mechanic's Magazine, No. 22, p. 344.

For, join DF, EF. Then the two triangles ADF, AEF, having the two sides AD, DF, equal to the two AE, EF (being equal radii), and the side AF common, they are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angle BAF equal to the angle CAF.

Scholium. In the same manner is an arc of a circle bisected.

PROBLEM III.

Ar a given point c, in a line AB, to erect a perpendicular. From the given point c, with any radius, cut off any equal parts CD, CE, of the given line; and, from the two centres D and E, with any one radius, describe arcs intersecting in F; then join CF, which will be perpendicular as required.

CEB

Then the two

For, draw the two equal radii DF, EF. triangles CDF, CEF, having the two sides CD, DF, equal to the two CE, EF, and cr common, are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the two adjacent angles at c equal to each other; therefore the line cr is perpendicular to AB (def. 11).

Otherwise.

WHEN the given point c is near the end of the line.

From any point D assumed above the line, as a centre, through the given point c describe a circle, cutting the given line at E; and through E and the centre D, draw the diameter EDF; then join cr, which will be the perpendicular required.

ΑΕ

CB

For the angle at c, being an angle in a semicircle, is a right angle, and therefore the line CF is a perpendicular (by def. 15).

PROBLEM 1V.

From a given point a, to let fall a perpendicular on a given

line BC.

From the given point a as a centre, with any convenient radius, describe an arc, cutting the given line at the two points D and E; and from the two centres D, E, with any radius, describe two arcs, intersecting at F; then draw AGF, which will be perpendicular to BC as required.

A

Then the

For, draw the equal radii AD, ae, and df, ef. two triangles ADF, AEF, having the two sides AD DF, equal to the two AE, EF, and AF common, are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angle DAG equal the angle EAG. Hence then, the two triangles ADG, AEG, having the two sides ad, ag, equal to the two AE, AG, and their included angles equal, are therefore equiangular (th. 1), and have the angles at & equal; consequently AG is perpendicular to BC (def. 11).

Otherwise.

WHEN the given point is nearly opposite the end of the line. From any point D, in the given line BC, as a centre, describe the arc of a circle through the given point A, cutting BC in E; and from the centre E, with the radius EA, describe another arc, cutting the former in F; then draw AGF, which will be perpendicular to Bc as required.

B

F

For, draw the equal radii DA, DF, and ea, ef. Then the two triangles DAE, DFE, will be mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angles at p equal. Hence, the two triangles DAG, DFG, having the two sides DA, DG, equal to the two DF, DG, and the included angles at D equal, have also the angles at G equal (th. 1); consequently those angles at G are right angles, and the line AG is perpendicular to DG.

PROBLEM V.

Ar a given point A, in a line AB, to make an angle equal to a given angle c.

Then,

From the centres a and c, with any one radius, describe the arcs DE, fg. with radius DE, and centre F, describe an arc, cutting FG in G. Through G draw the line AG, and it will form the angle required.

equal lines or radii,

For, conceive the DE, FG, to be drawn.

Then the two trian

gles cde, afg, being mutually equilateral, are mutually equiangular (th. 5), and have the angle at a equal to the angle c.

PROBLEM VI.

THROUGH a given point a, to draw a line parallel to a given

A

line BC.

From the given point a draw a line Ad to any point in the given line BC.

Then

draw the line EAF making the angle at a equal to the angle at D (by prob. 5); so shall EF be parallel to BC as required.

For, the angle D being equal to the alternate angle a, the lines BC, EF, are parallel, by th. 13.

PROBLEM VII.

To divide a line AB into any proposed number of equal

parts.

Draw any other line Ac, forming any angle with the given line AB; on which set off as many of any equal parts AD, DE, EF, FC, as the line AB is to be divided into. Join BC; parallel to which draw the other lines FG, EH, DI: then these will divide AB

E
D

AIHGB

in the manner as required.-For those parallel lines divide both the sides AB, AC, proportionally, by th. 82.

PROBLEM VIII.

To find a third proportional to two given lines AB, ac.

Place the two given lines AB, ac, forming any angle at A; and in AB take also AD equal to AC. Join BC, and draw DE parallel to it; so will aɛ be the third proportional sought.

For, because of the parallels, BC, de,

the two lines AB, AC, are cut propor.

A

A

-B C

E

tionally (th. 82); so that AB AC AD or AC AE; there

fore AE is the third proportional to AB, AC.

PROBLEM IX.

To find a fourth proportional to three lines ab, ac, ad. Place two of the given lines AB, AC, making any angle at A; also place AD on AB. Join BC; and parallel to it draw

To find a mean proportional between two lines ab, BC.

Place AB, BC, joined in one straight line AC on which, as a diameter, describe the semicircle ADC; to meet which erect the perpendicular BD; and it will be the mean proportional sought, between AB and BC (by cor. th. 87).

A

-B

D-C

A

ов с

PROBLEM XI.

To find the centre of a circle.

Draw any chord AB; and bisect it perpendicularly with the line CD, which will be a diameter (th. 41, cor.). Therefore CD bisected in o, will give the centre, as requir

ed.

PROBLEM XII.

B

To describe the circumference of a circle through three given

points A, B, C.

From the middle point в draw chords BA, BC, to the two other points, and bisect these chords perpendicularly by lines meeting in o, which will be the centre. Then from the centre o, at the distance of any one of the points, as oa, describe a circle, and it will pass through the two other points B, C, as required.

B

C

For the two right-angled triangles OAD, OBD, having the sides AD, DB, equal (by constr.), and oD common, with the included right angles at D equal, have their third sides oa, OB, also equal (th. 1). And in like manner it is shown that oc is equal to oв or oa. So that all the three OA, OB, oc, being equal, will be radii of the same circle.