PROBLRM 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 centre 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.

PROBLEM Xil

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 semircle, is a right angle, and consequently AD is perpendicular to the radius DO, or is a tangent to the circle (th. 46.)

PROBLEM XXIII,

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

Ε

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.

B

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.

B

E

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.

PROBLEM XXVI.

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.

For, since the angle DAE is equal to the angle DAG, and the angles at E, G, right angles (by con

struction), the two triangles, ADE, ADG, are equiangular; and, having also the side AD common, they are identical, and have the sides DE, DG, 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.

PROBLEM XXVII.

To inscribe a square in a given circle.

Draw two diameters AC, BD, crossing at right angles in the centre E. Then join the four extremities A, B, C, D, with right lines, and these will form the inscribed square ABCD.

For the four right-angled triangles AEB, BEC, CED, DEA, are identical, because they have the sides EA, EB, EC, ED, all equal, being radii of the circle, and the four included angles at E all equal, being right angles, by the construction. Therefore, all their third sides AB, BC, CD, DA, are equal to one another, and the figure ABCD is equilateral. Also, all its four angles, A, B, C, D, are right ones, being angles in a semicircle. Consequently the figure is a square.

PROBLEM XXVIII.

To describe a square about a given circle.

Draw two diameters AC, BD, crossing at right angles, in the centre E Then through their four extremities draw FG, IH, parallel to AC, and FI, GH, parallel to BD, and they will form the square FGHI,

For, the opposite sides of parallelograms being equal, FG and IH are each equal to the diameter AC, FI and GH each equal to the diameter BD; so that the figure is equilateral. Again, because the opposite angles of parallelograms are equal, all the four angles F, G, H, I, are right angles, being equal to the opposite angles at E. So that the figure FGHI, having its sides equal, and its angles right ones, is a square, and its sides touch the circle at the four points A B, C, D, being perpendicular to the radii drawn to those points.

PROBLEM XXIX.

To inscribe a circle in a given square.

Bisect the two sides FI, FG, in the points A and B (last fig.) Then, through these two points draw AC parallel to FG or IH, and BD parailed to FI or GH. Then the point of intersection E will be the centre, and the four lines EA, EB, EC, ED, radii of the inscribed circle.

For, because the four parallelograms EF, EG, EH, EI, have their opposite sides and angles equal, therefore all the four lines EA, EB, EC, ED, are equal, being each equal to half a side of the square. So that a circle described from the centre E, with the distance EA, will pass through all the points A, B, C, D, and will be inscribed in the square, or will touc's its four sides in those points, because the angles there are right ones.

PROBLEM XXX.

To describe a circle about a given square.

(See fig. Prob. xxvii.)

Draw the diagonals AC, BD, and their intersection E will be the centre.

As the diagonals of a square bisect each other (th. 40), then will EA, EB, EC, ED, be all equal, and consequently these are radii of a circle passing through the four points A, B, C, D.

PROBLEM XXXI.

To find a third-proportional to two given lines, AB, AC. Place the two given lines AB, AC (or two lines equal to them), to form any angle at A; and in AB set off AD = AC. Join BC, and draw DE parallel to it; so will AE, on the line AC, be the third proportional sought.

For, since DE is parallel to BC, the two lines AB, AC, are cut proportionally by DE (th. 82): hence,

A

-B

AB: AC :: AD (= AC) : AE, and AE is, therefore, the third proportional required.

PROBLEM XXXII.

To find a fourth proportional to three given lines, AB, AC, AD.

A

Place two of the given lines AB, AC, or their equals, to make any angle at A; and on AB set off, or place, Athe other line AD, or its equal. Join BC, and paral- A lel to it draw DE: so shall AE be the fourth proportional as required.

For, because of the parallels BC, DE, the two sides

D

-C

AB, AC, are cut proportionally (th. 82): so that AB: AC :: AD: AE

PROBLEM XXXIII.

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-C

PROBLEM XXXIV.

To divide a given line in extreme and mean ratio.

Let AB be the given line to be divided in extreme and mean ratio, that is, so that the whole line may be to the greater part, as the greater is to the less part.

Draw BC perpendicular to AB, and equal to half AB. Join AC; and with centre C and distance CB, describe the circle BD; then with centre A and distance AD, describe the arc DE; so shall AB be divided in E in ex. treme and mean ratio, or so that AB: AE:: AE: EB.

Produce AC to the circumference at F. Then, ALF

ов

B

B

being a secant, and AB a tangent, because B is a right angle: therefore the rectangle AF AD is equal to AB (cor. 1, th. 61); consequently the means and extremes of these are proportional (th. 77), viz. AB: AF or AD + DF :: AD: AB. But AE is equal to AD by construction, and AB = 2 BC= DF; therefore, AB AE + AB :: AE: AB or AE + EB; and by division, AB: AE:: AE: EB.

PROBLEM XXXV.

To cut a given line AB in a point F, so that the square of the one part RF may be equal to the rectangle of the whole line AB and the other part AF.

Produce AB till BC be equal to it, erect the perpendicular BD equal to AB or BC, bisect BC in E, join ED and make EF equal to it; the square of the segment BF is equivalent to the rectangle contained by the whole BA and its remaining segment AF. The line AB is then said to be divided by medial section at the point F.

G

H

K

B

E

For on BC construct the square BG, make BH equal to BF, and draw IHK and FI parallel to AC and BD. Since AB is equal to BD, and BF to BH; the remainder AF is equal to HD: and it is further evident, that FH is a square, and IC and DK are rectangles. But BC being bisected in E and produced to F, the rectangle under CF, FB, or the rectangle IC, together with the square of BE, is equivalent to the square of EF or DE. But the square of DE is equivalent to the squares of DB and BE; whence the rectangle 1C, with the square of BE, is equivalent to the squares of DB and BE; or, omitting the common square of BE, the rectangle IC is to the square of DB. Take away from both the rectangle BK, and there remains the square BI, or the square of BF, = to the rectangle HG, or the rectangle contained by BA and AF.

Cor. Hence also the construction of another problem of the same nature; in which it is required to produce a straight line AB, such that the rectangle contained by the whole line thus produced and the part produced, shall be equivalent to the square of the line AB itself.

Bisect AB in C, draw the perpendicular BD = BC, join AD and continue it until DE = DB or BC, and on AB produced take AF AE: the line AF is the required extension of AB. For make DG = DB or BC; and because the rectangle EA, AG together with the square of DG or DB, is equivalent to the square of DA or to the squares of AB and DB; the rectangle EA, AG, or FA, AC, is equivalent to the square of AB.

E

D

G

C

B

PROBLEM XXXVI,

Given either one of the sides AB, or the base a b, to construct an isosceles triangle, so that each of the angles at the base may be double of its vertical angle.

First, let one of the sides AB be given. By the last problem divide it into two parts, AC, CB, such that CB2 AB X AC. Construct the triangle, having the base = CB, and each of the two sides = AB.

C

A

B