Suppose that AL, BL are the lines required to be drawn, and ALB to the given angle, join AB, and describe through the points A, L, a circle: then it is evident that the segment ALB is one capable of containing the given angle.

Therefore join AB, on it describe a segment of a circle capable of containing the given angle, either of the points where its circumference cuts the given line, is the point to which the lines are to be drawn.

PROP. 15. THEOR.

If a right line (CD) be divided into five equal parts, the square of the line (CB) made up of four of those parts, plus the square of the line (CA) made up of three of them is equal to the square of the whole

line.

Fig. 27.

2

For since CB is divided in A and AB to BD, (prop. 8, b. 2, Elr.) the 2 of CD is to CA2 + 4 CB X BA, but 4 CB × BA is to CB, .'. &c.

PROP. 16, PROB.

Given the hypothenuse to describe a right angled triangle, so that the hypothenuse and one side may be together double of the third side.

Divide the given right line into five parts, four of those parts shall be to the greater side, and three to the lesser.

For by adding the lesser to the given line their sum shall be to 8 of the parts, which is .. double of the other.

Cor. And those lines shall form a right angled triangle.

2

For the of the given line is to the sum of their rs, by the foregoing prop.

PROP. 17. PROB.

Given the hypothenuse of a right angled triangle, and the sum or difference of the base and perpendicular, to construct the triangle.

Fig. 28.

=

ANALYSIS.

Suppose it done, and that ABC is the required triangle: let AC be the given hypothenuse, and AD; AE to the sum or difference of the base and perpendicular; BD or BE is to BC, join CE,CD, then ĈED is evidently an isosceles right angled triangle, ... each of the angles CED,CDE is half a right angle, .. the points D or E being given, the right line EC or DC is given in position.

If therefore a right line DC or EC be drawn from either extremity of the sum or difference, making the angle CED or CDE = to half a right angle, and inflect from the other extremity A a right line AC (= to the hypothenuse) to meet EC or DC, and let fall from the point C in which they meet a perpendicular CB, ACB is evidently the required triangle.

PROP. 18, PROB.

To draw through a given point, (L) either within or without a given circle (ABC), a right line, the part of which intercepted by the circle, shall be equal to a given right line, not greater than the diameter of the circle.

Suppose it done, and that LD is the line required to be drawn; let AB be the given line, draw from the centre I, IE and IF at right angles to them, IE, IF are =,.. the circle described from I as centre, with IE as radius, will touch both lines in E and F.

Therefore let fall IE at right angles to the given line, with I as centre and IE as radius, describe a circle ETF; draw through the given point L a tangent LD to this circle the part of it CD is evidently = to AB.

PROP. 19, THEOR.

The squares of the diagonals of any parallelogram, are together equal to the sum of the squares of the sides.

Fig. 21.

For the diagonals evidently bisect one another, and the 2rs of one pair of sides (AB,BC) are together to 2 AE + 2 EB', and the 2rs of the other pair of sides are together to 2 AE2 + 2 ED', but 4 AE +4 EB' are together to AB + BD,. &c.

=

==

PROP. 20, THEOR.

If in a circle two chords (AB, CD) cut one another at right angles the sum of the squares of the segments› is equal to the square of the diamater.

Fig. 30.

2

Draw the diamater BF; join CF, FA and AD; Then FB is to FC2+ CB2 (prop. 47. 1. Elr.) but CB is to CH + HB, and CF is evidently =to AD*, but AD is to AH2+ HD2.. FB is to CH+HD+BH+HA.

CF is to AD for CD is parallel to FA. join CA and then it is evident.

From this also it is evident that in a circle parallel chords intercept equal arches.

PROP. 21, THEOR.

If from the centre (A) of a given circle a right line be drawn to any point (C) in a chord the square of this drawn line together with the rectangle under the segments of the chord is equal to a square of the radius.

Fig. 31.

Produce AC both ways to L, F.

Then the rectangles LCF, DCB are but the rect. LCF with AC is to AF.. the rect. DCB with AC2 = to AF2.

is

PROP. 22, THEOR.

If in a circle two chords (DB, EI) cut a diameter (LF) at the same point and at equal angles they are equal to one another.

Fig. 31.

Bisect DB and EI in K, H join A K and AH. Then, because the angles ACH, poth.) and the angles AKC, AHC and AC common AK is to AH, ... DB is to IE (prop. b. 3 Elr).

ACK are (by hy(by prop. b. 3 Elr.) (prop, 26. b.1. Elr.)

PROP. 23, PROB.

The diameter (AC) of a circle being produced to a given point (B) to find in the produced part a point from which if a tangent be drawn to the circle it shall be equal to the segment of the produced part between this point and the given point.

Suppose E is the required point, and EF the tangent join BF and produce it to L, join L with the centre O and join OF.

Then the angles LFO, OFE. EFB are evidently together to the angles BLO, LOB, OBL; and LFO is to FLO and also EFB to EBF .. EFO is LOB... LOB is a right angle.

=

= to

Then draw LO at right angles to BA join BL and FO and draw FB at right angles to FO; E is evidently the required point.

For the angles at F are together to the angles at B, F, O but ŎFL is = to OLF and EFO to BOL .. the angle EBF is to EFB .. &c. &c.

PROP. 24. PROB.

To describe a circle which shall have a given radius, and its centre in a given right line (EF) and shall touch another right line (BF) making a given angle with the former.

Fig. 33.

ANALYSIS.

Suppose it done, and that CL is the circle required to be described, join D the centre and point of contact C, CD is to the given radius and is at right angles to the given line BF.

Then inflect between BF and EF a right line CD = to the given radius, and making at the point C in which it meets B a right angle (prop. 27. on b. 1).

It is evident that the circle described from D as a centre with DC as radius touches BF (in C) has the given radius and its centre in EF (viz. D).

PROP. 25, THEOR.

the diagonals (IL BD) of a quadrilateral figure inscribed in a circle cut one another at right angles, the sum of the squares of the opposite sides are equal. Fig. 34.

to the

rs of BF,

'rs of Bb

For the 'rs of BL, ID are FL, DF, FI (prop. 47. b. 1. Elr.) also the LD are together to them.. BL+DI is = te BI2 + LD2.

PROP. 26, THEOR.

If two points be taken in the diamiter of a circle equally distant from the centre, the sum of the squares of the lines drawn from these to any point in the circumference is always the same.

2

For the sum of their rs is a of the distance from the given points.

2

to a☐ of the radius centre to

PROP. 27, THEOR.

either of the

*If from any two points (B, D) in the circumference of the greater of two concentric circles, two right lines (BA, DE,) be drawn to touch the other circle, they shall be equal to one another.