will bisect the angles, also produce DE until EF is = to it and draw FC; then the triangles ADE and FEC are evidently and have also their sides and angles respectively to one another, and also the triangles CDE and CEF are in every way, .. the angle FCD is the angle of an equilateral triangle, and since the angles FDC,DFC are they must be the angles of an equilat... DF is to FC or DA, but DF is double of DE or DI, ... DA is double of DI or DE,,'. AI is three times DI or DE, .. &c,

PROP. 4, PROB.

Given of any triangle the base, line bisecting the base, and sum of the sides, to find the triangle.

Take from the 2 of the sum two rs of half the base, plus two'rs of the bisecting line, take half the remainder and find the side of a2 to it, then cut the given sum so that the rectangle under its segments may be to this, &c,

PROP. 5, THEOR.

If a tangent (FH) be drawn to a circle, parallel to a chord (AB), the point of contact (F) will be the middle point of the arch cut off by that chord.

Suppose it to be the case; join FA,FB; then since F is the middle point of the arch FA,FB are,.. the angles FAB, FBA are; if this can be proved the proposition is true.

to

Since FH is parallel to AB the angle ABF is BFH, but BFH is to BAF, (prop. 32, b. 3,) .. ABF is to BAF, .. FA is to FB, &c.

PROP, 6, PROB.

To draw a common tangent, to two given circles, in a transverse direction.

Fig. 17.

ANALYSIS.

Suppose it done, and that OL is the required tangent, join FO and LE, (i. e. the centres with the points of contact) produce FO 'till OH is to LE, join

HE, because OL is a tangent the angles FOL, ELO are right,.. FH is parallel to LE, and since OH is = to LE, HE is parallel to OL, .. the angle FHE is a right angle.

Then join the centres F, E; and on FE describe a semicircle FHE, in it inflect a right line FH to the sum of the radii FO, LE, join HE, draw EL parallel to HF and join LO: LO is the tangent; the demonstration is evident.

Also a direct tangent can be drawn to two given circles.

1. If the given circles be connect their centres, draw a radius in each on the same side and at right angles to the connecting line, and connect their extremities at the circumference; this connecting line is the required tangent.

For the circles being the radii are, and they being and at right angles to the line connecting the diameter, they are parallel,.. the connecting lines are and par. .. the radii are at right angles to the one at the circumferences, .. it is a tangent.

2. If the given circles be unequal

Suppose IH is the required tangent, TI and HE are evidently par. to one another, make IF = to HE and join FE, it is and par. to IH,.. the angle IFE is right and IF is the difference between TI and HE; with FI a radius describe a circle FA, EF is evidently a tangent to it.

Therefore with the centre I and the difference of the radii IF describe a circle FA, and from the centre E of the lesser circle draw a tangent EF to it, produce IF to T, draw EH par. to IT and join TH; it is evi dently the required tangent.

PROP. 7. THEOR.

The distance of the chord (AB), whose square is threefourths of the square of the diameter from the centre (CD,) is half the radius.

Fig. 19.

Bisect AB in F, and join EF, EA and EB

AB is

13

2

of CD2, but AB is to four times AE four times EF2, ... four times ED which is of CD2, .:. EF is

EF must be

to ED.

PROP. 8. PROB.

to

Given of any triangle, the base, vertical angle and difference of the sides, to find it.

Suppose it done, and that ADC is the required triangle, let AC be to the given base, AB to given difference and ADC to the given angle; join BC. Then because AB is the difference between AD and CD, DB and DC are =, the angles DBC,DCB are =, DBC is half the supplement of the given angle and .. ABC is the supplement of half the supplement of the given angle.

Therefore, on the given base describe a segment of a circle capable of containing the given angle, and also a segment capable of containing the supplement of half the supplement of the given angle; inflect in the latter segment, from one extremity of the given base, a right line to the given difference, produce it to meet the periphery of the other segment, and join its extremity with the other extremity of the given base; thus is formed the required triangle. The demonstration is evident.

PROP. 9, THEOR.

If from any point (F) in a rectangle (ABCD), four right lines be drawn to the four angles, the sums of the squares of those drawn to the opposite angles, will be equal, (i. e. FA2 + FC2 = FB2 + FD3). Fig. 21.

For let the diagonals AC and BD be drawn; they bisect each other in E, and join EF, then the triangles ABC,BAD being in every way, (prop. 24, b. 1,) thence will A E, i. e. AC DE, i. e. DB, but FA FC2 2 AE2 (DE2) +2 EF2 FB FD2.

PROP. 10. PROB.

Given the base (AB) and sum of the squares of the sides of any triangle, to find the locus of the

Suppose it done, and that AEB,ADB are any num ber of triangles on the given base, the sum of the rs of their sides are, as being each = to the given quantity; then it is evident that the difference between the sum of the rs of AE,EB and two 'rs of half AB is the 2 of the bisecting line EC, also the difference between the sum of the rs of AD,DB and two r of half AB is to the 2 of the bisecting line DC, .. DC is to EC',... CE is to CD, .. the circle described with C as a centre and CE as radius, will pass through the vertices of any number of triangles on the given base, the sum of 2rs of whose sides shall be =

Therefore, from the sum of the ers take two 2rs of AB, the remainder shall be to two rs of the bisecting line, then the bisecting line can be found; and it is evident that the required locus lies in the circumference of the circle of which this besecting line is radius.

PROP. 11. THEOR,

In every quadrilateral figure (ABCD) the sum of the squares of the sides is equal to the sum of the squares of the diagonals, plus four times the square of the line (EF) joining the points of bisection of the diagonals.

Fig. 23.

Join EA, ED.

Then the sum of the ers of CA,AB is

2rs

to two

Ders of CE + two rs of EA, and the 2rs of CD, DB are to two 2rs of CE+ two there are four 2rs of CE which are together and there are also two 2rs of AE and two and those are together to AD2 + four prop. 13. b. 3.

of ED, or to CB2, 2rs of DE, times FE.

PROP. 12. THEOR.

If any number of equal right lines (EF IL KT) be placed in a circle, the locus of their points of bisection will be in the circumference of a circle.

Fig. 24.

For if right lines OK,OP be drawn from the centre O to the points of bisection K and P, they are evidently to one another, (prop. 14, b. 3, Elr.).. the circle described with O as a centre and OK as radius shall pass through the points of bisection of every right line placed within that circle.

PROP. 13, THEOR.

The locus of the centres of any number of circles which shall touch two given right lines (AD,AB) that are not parallel, is the right line (AF) bisecting the angle (BAD) made by the given lines.

Fig. 25.

For, from any points I T in AF let fall perpendiculars IB,ID or TL,TS; it is evident that IB, ID are to one another, and .. the circle described with the centre I, and radius ID, touches the given lines: also the circle described from T as a centre and IL as radius, touches the given lines, .'. &c.

Cor. From this proposition, by being given the radius, a circle can be described to touch two given lines, that are not parallel.

For, draw AP at right angles to AB, and make it = to the given radius, draw PI parallel to AB and draw IB par. to PA; IB is evidently to the given radius, ... &c.

PROP. 14, PROB.

To draw from two given points (A, B) on the same side of a given right line (CD,) two right lines, to the same point in the given line, which shall contain a given angle.