For, the angle D being equal to the alternate angle A, the lines BC, EF,

are parallel, by th. 13.

キ

PROBLEM IX,

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

A

G

PROBLEM X.

To make a square on a given line AB.

Raise AD, BC, each perpendicular and equal to AB; and join DC: so shall ABCD be the square sought.

D

B

For all the three sides AB, AD, BC, are equal, by the construction, and DC is equal and parallel to AB (by th. 24); so that all the four sides are equal, and the opposite ones are parallel. Again, the angle A or B, of the parallelogram, being a right angle, the angles are all right ones (cor. 1, th. 22). Hence, then, the figure, having all its sides equal, and all its angles right, is a square (def. 34).

PROBLEM XI.

To make a rectangle, or a parallelogram, of a given length and breadth, AB, BC.

Erect AD, BC, perpendicular to AB, and each equal to BC; then join DC, and it is done.

The demonstration is the same as the last problem. And in the same manner is described any oblique parallelogram, only drawing AD and BC to make the given oblique angle with AB, instead of perpendicular to it.

D

B.

To make a rectangle equal to a given triangle ABC. Bisect the base AB in D; then raise DE and BF perpendicular to AB, and meeting CF parallel to AB, at E and F; so shall DF be the rectangle equal to the given triangle ABC (by cor. 2. th. 96\

CE

D

A

To make a square equal to the sum of two or more given squares. Let AB and AC be the sides of two given squares. Draw two indefinite lines AP, AQ, at right angles to each other; in which place the sides AB, AC, of the given squares; join BC: then a square described on BC will be equal to the sum of the two squares described on AB and AC (th. 34).

A

A

P

B

D

IQ BE

In the same manner, a square may be made equal to the sum of three or more given squares. For, if AB, AC, AD, be taken as the sides of the given squares, then, making AE= BC, AD = AD, and drawing DE, it is evident that the square on DE will be equal to the sum of the three squares on AB, AC, AD. more squares.

And so on for

PROBLEM XIV.

To make a square equal to the difference of two given squares. Let AB and AC, taken in the same straight line, be equal to the sides of the two given squares. From the centre A, with the distance AB, describe a circle; and make CD perpendicular to AB, meeting the circumference in D so shall a square described on CD be equal to AD? - AC2, or AB2—— AC2, as required (cor. th. 34).

-

A CB

PROBLEM XV.

To make a triangle equal to a given quadrilateral ABCD. Draw the diagonal AC, and parallel to it DE, meeting BA produced at E, and join CE; then will the triangle CEB be equal to the given quadrilateral ABCD.

For, the two triangles ACE, ACD, being on the same base AC, and between the same parallels AC, DE, are equal (th. 25); therefore, if ABC be added to each, it will make BCE equal to ABCD (ax. 2).

A

B

To make a triangle equal to a given pentagon ABCDE. Draw DA and DB, and also EF, CG, parallel to them, meeting AB produced at F and G; then draw DF and DG; so shall the triangle DFG be equal to the given pentagon ABCDE.

For, the triangle DFA = DEA, and the triangle DGB = DCB (th. 25); therefore, by adding DAB to the equals, the sums are equal (ax. 2), that is, DAB

E

A

+ DAF + DBG = DAB+ DAE + DBC, or the triangle DFG to the pentagon ABCDE.

PROBLEM XVII.

To make a square equal to a given rectangle ABCD

Produce one side AB, till BE be equal to the other side BC. On AE as a diameter describe a circle meeting BC produced at F: then will BF be the side of the square BFGH, equal to the given rectangle BD, as required; as appears by cor. th. 87, and th. 77.

D

A

G

H

E

PROBLEM XVIII.

To describe a circle about a given triangle ABC.
Bisect any two sides with two of the perpendiculars
DE, DF, DG, and D will be the centre.

E

B

For, Join DA, DB, DC. Then the two right-angled triangles DAF, DBE, have the two sides, DE, EA, equal to the two DE, EB, and the included angles at E equal: these two triangles are therefore identical (th. 1), and have the side DA equal to DB. In like manner it is shown, that DC is also equal to DA or DB. So that all the three, DA, DB, DC, being equal, they are radii of a circle passing through A, B, and C.

Note. The problem is the same in effect when it is required—

To describe the circumference of a circle through three given points A, B, C.

Then, from the middle point B draw chords BA, BC,

to the two other points, and bisect these chords perpendicularly by lines meeting in O, which will be the centre. Again, 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. The demonstration is evidently as above.

PROBLEM XIX.

An isosceles triangle ABC being given, to describe another on the same base AB, whose vertical angle shall be only half the vertical angle C.

From C as a centre, with the distance CA, describe the circle ABE. Bisect AB in D, join DC, and produce to the circumference E, join EA and EB, and ABE shall be the isosceles triangle required.

For, since in the triangle EDA, EDB, AD is equal to DB, and DE common to both, and the right angle EDA, equal to the right angle EDB, the side EA must be equal to the side EB, the tri

angle ↑ EB, is therefore isosceles, and the angle

E

B

D

ACB at the centre, must be double of the angle AEB at the circumference

for they both stand on the same segment AB.

PROBLEM XX.

Given an isosceles triangle AEB, to erect another on the same base AB, which shall have double the vertical angle E.

Describe a circle about the triangle AEB, find its contre C, and join CA, CB, and ACB is the triangle required.

The angle C at the centre is double of the angle E at the circumference, and the triangle ACB is isosceles for the sides CA, CB being radii of the same circle are equal.

PROBLEM XXI.

To find the centre of a given circle.

Draw any chord AB; and bisect it perpendicularly with the line CD: this (th. 41, cor.) will be a diameter. Therefore bisect CD in O, which will be the centre, as required.

To draw a tangent to a circle, through a given point A.

1. When the given point A is in the circumference of the circle: join A and the centre O; perpendicular to which draw BAC, and it will be the tangent, by th. 46.

2. When the given point A is out of the circle: draw AO to the centre O; on which as a diameter describe a semicircle, cutting the given circumference in D; through which draw BADC, which will be the tangent as required.

For, join DO. Then the angle ADO, in a semicircle, is a right angle, and consequently AD is perpendicular to the radius DO, or is a tangent to the circle (th. 46.)

PROBLEM XXIII.

B

On a given line AB to describe a segment of a circle, to contain a given angle C

At the ends of the given line make angles DAB, DBA, each equal to the given angle C. Then draw AE, BE, perpendicular to AD, BD; and with the centre E, and radius EA or EB, describe a circle; so shall AFB be the segment required, as any angle F made in it will be equal to the given angle C.

For, the two lines AD, BD, being perpendicular to the radii EA, EB (by construction), are tangents to

A

the circle (th. 46); and the angle A or B, which is equal to the given angle C by construction, is equal to the angle F in the alternate segment AEB (th. 53).

PROBLEM XXIV.

To cut off a segment from a circle, that shall contain a given angle C

Draw any tangent AB to the given circle; and a chord AD to make the angle DAB equal to the given angle C; then DEA will be the segment required, any angle E made in it being equal to the given angle C.

PROBLEM XXV.

To inscribe an equilateral triangle in a given circle.

Through the centre C draw any diameter AB. From the point B as a centre, with the radius BC of the given circle, describe an arc DCE. Join AD, AE, DE, and ADE is the equilateral sought.

D

B

Join DB, DC, EB, EC. Then DCB is an equilateral triangle, having each side equal to the radius of the given circle. In like manner, BCE is an equilateral triangle. But the angle ADE is equal to the angle ABE or CBE, standing on the same arc AE; also the angle AED is equal to the angle CBD, on the same arc AD; hence the triangle DAE has two of its angles, ADE, AED, equal to the angles of an equilateral triangle, and therefore the third angle at A is also equal to the same; so that the triangle is equiangular, and therefore equilateral.

To inscribe a circle in a given triangle ABC.

Bisect any two angles A and B, with the two lines AD, BD. From the intersection D, which will be the centre of the circle, draw the perpendiculars DE, DF, DG, and they will be the radii of the circle required.

A

For, since the angle DAE is equal to the angle DAG, and the angles at E, G, right angles (by construction), the two triangles, ADE, ADG, are equiangular; and, having also the side AD common, they are identical, and have the sides DE, IG, equal (th. 2). In like manner it is shown, that DF is equal to DE or DG.

Therefore, if with the centre D, and distance DE, a circle be described, it will pass through all the three points, E, F, G, in which points also it will touch the three sides of the triangle (th. 46), because the radii DE, DF DG, are perpendicular to them.