PROBLEM S.

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 cp, which will bisect the given line AB in the point E.

For, draw the radii Ac, EC, ad, bd. Then, because all these four radii are equal, and the side co common, the two

triangles 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 an equal to EB. Q. E. D.

PROBLEM II.

To Bisect an Angle BAC.

PROM the centre A, with any radius, describe an arc, cutting off the equal lines AD, AE; and from the two centres D, E, with the same radius, describe arcs intersecting in F; then draw AF, which will bisect the angle A as required.

For, join DF, EF. Then the two triangles ADF, AEF, having the two sides AD,

D

B

pr, 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 car.

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

PROBLEM III.

At a Given Point c, in a Line AB, to Erect a Perpendicular.

FROM the given point c, with any radius, cut of 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.

AD CEB

Then the two tri

For, draw the two equal radii DF, EF. angles 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 CF, which will be the perpendicular required.

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

PROBLEM IV.

ΑΕ

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.

EF.

For, draw the equal radii ad, aɛ, and DF,

B

A

G

Then the two triangles ADF, AEF, having the two sides AD, DF, equal to the two AE, EF, and AF common, are mutu

ally

ally 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 G equal; consequently AG is perpendicular to ac (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 в; 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

D

C GE

F

Then the

For, draw the equal radii da, df, and EA, EF. two triangles DAE, DFE, will be mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angles at D» equal. Hence, the two triangles DAG, dfg, having the two sides DA, DG, equal to the two DF, DG, and the included angles at n equal, have also the angles at & equal (th. 1); consequently those angles at a are right angles, and the line AG is perpendicular to DG.

PROBLEM V.

At a Given Point A, in a Line AB, to make an Angle Equal to a Given Angle c.

Then,

FROM the centres ▲ 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 & draw the line AG, and it will form the angle required.

equal lines or radii,
Then the two trian-

A

E

D

FB

For, conceive the DE, FG, to be drawn. gles CDE, AFG, being mutually equilateral, are mutually equiangular (th. 5), and have the angle at ▲ equal to the angle c.

PROBLEM

PROBLEM VL

Through a Given Point A, to draw a Line Parallel to a Giver

Line BC.

Then

From the given point a draw a line AD to any point in the given line Bc. 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 ▲, the lines BC, EF, are parallel, by th. 13.

FROBLEM VIL

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

AIHGB

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

PROBLEM VIII.

To find a Third Proportional to Two given Lines AB, AÇI

PLACE the two given lines AB, AC,

and in AB take

forming any angle at a ;
also AD equal to AC.
draw DE parallel to it;
the third proportional sought.

A

Join BC,
so will aɛ be

B

For, because of the parallels BC, DE, the two lines AB, AC, are cut proportionally (th. 82); so that AB AC 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 Ab on AB. Join BC; and parallel to it draw DE :

BC.

A-
D-C

-B

To find a Mean Proportional between Two Lines ab, 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).

PROBLEM XI.

To find the Centre of a Circle.

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

OB C

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.

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 OB or oд. So that all the three os, oв, oc, being equal, will be radii of the same circle.

PROBLEM