... the angle XNY, and the angle NSY the difference of the angles XIZ, and ZIO the difference of the base angles and is. given; join IY and YD, then IY 10 ID... the angle NDY is given, and XY is given, also DX; thence we can derive the following

construction.

=

=

On XY describe a segment to of a circle to contain an angle XDY, inflect a chord XD join DY and make an angle DYI IDY make IN IY let fall IZ at right angles, &c.

=

PROP. 65. THEOR.

Of all the right lined figures, contained under the same number of sides, and inscribed in the same circle, that is the greatest whose sides are all equal.

Fig. 66.

For if possible let some polygon ABCFE whose sides CF FE are unequal be the greatest.

Let CDE be an isosceles triangle in the same segment with CFE; which being greater than CFE (prop. on b 3.) the whole polygon ABCDE will be greater than ABCFE, which is absurd,.. the polygon is greatest when all the sides are .

PROP. 66. PROB.

To describe two circles, each having a given radius which shall touch a given right line [DE] both on the same side and shall also touch one another.

Suppose the circles CI and FI touch the given right line DE in C and F and each other in I: join the centres L,O and also the centres and points of contact LC and OF, and draw LH par. to DE: then it is evident that OF is at right angle to DE; .. that HF is = to LC or LI and also that LO is = to the sum of the given radii.

Then in DE take any point F draw FG at right angles to DE and to either of the given radii; make FH to the other, draw HL parallel to DE and from O infleet OL to the sum of the given radii, then the circles described from O and L as the centres with the given radii shall evidently touch the given line and one another.

PROP. 67. THEOR.

If from a given point (C) in the circumference of a circle, two right lines (CA CB) be drawn to the extremeties of a given chord (AB), the angle (CGK) which one of them makes with (KD) any perpendi cular to the chord, shall be equal to the angle (ACF) which the other makes with a diameter (CF) of the circle which passes through the given points; and also the angle (CEK) made by one of the drawn lines (CA) produced, and the perpendicular produced is equal to the angle (BCF) contained by the other drawn line and diameter.

Fig. 68.

For the angle CAK is to the angles AEK AKE but the angle AKE is ev dently AKB and ACB is also AKB (prop. 20. b. 3 Elr) .. CAK is to CEK + ECG, but CAK is to ACK, if ... ECB is taken from both, BCF will remain to KEC.

Again, the angle CGE is to the angle GCE + GEC, but BCF is to GEC .. EGB or CGK is = to ECB BCF or to ACF.

PROP. 68. THEOR.

A circle cannot be described about any parallelogram which is not rectangular.

For the opposite angles of a pallelogram are to one another, and if a circle could be described, they would be to two right angles .. since they are = each of them should be a right angle, but they are not so by hypothesis, .. a circle cannot be described.

PROP 69, THEOR.

If two chords (AE CB) of a given circle, intersect one another in (I) the angle (CIE) of their inclination is equal to half the angle at the centre standing on the sum or differences of the arches intercepted between them, according as they meet within or without the given circle.

Fig. 69.

1st. Let them meet within the circle: through B draw BF parallel to AE: then the arch EF is — to AB

CEF is to the sum of CE AB: since BF is par. to AE, the angle CIE is to the angle CBF, but CBF is to half the angle at the centre standing on the arch CEF, .'. &c.

2d. If the two CA EO meet outside the circle: draw AH parallel to LE, then the arch AO is = to HE.. the arch CH is the difference between AO and CE, and since AH is parallel to LE, the angle CAH is — to to CLB,.. &e.

PROP. 70. THEOR.

If from any point (E) in the diameter of a eircle there be drawn two right line (EA EF) to the circumference, one to the bisection of a semicircular arch and the other at right angles to the diameter, the squares of those two lines are together equal to two squares of the radius.

Fig. 70.

2

Bisect the diameter in I, join TA IF: then since OA is to OL, the angles OIA LIA are and .. right: ... the of EA is to the of IA + the of IE, to each add the 2 of EF, the EA2 + is to IA2 + IE EF but IE2 + EF is to IF... EA2 + EF2 IA2+ IF..

=

PROP. 71, THEOR.

If two circles: (ACD ECF) touch one another, any right line (ACF) drawn through the point of contacts will cut off similar segments.

Fig. 71.

Draw DCE passing through the centre of one; and point of contact it shall pass also pass throngh the centre of the other, join DA and EF.

Then in the triangles DCA FCE the angles DCA FCE are, (prop. 15. b. 1. Elr.) and the angles DAC EFC are, being angles in semicircles; .. the angles ADC CEF are,.. the segments ADC CEF are similar: in like manner the segments AIC FLC can be proved similar.

PROP. 72. THEOR.

If two circles (ACD ECF) touch one another, two right lines (ACF DCE) drawn through the point of contact intercept arches, (DA FE) the chords of which are parallel.

Fig. 71.

For by the foregoing proposition the arches ADC CEF are similar, .. the angles ADC CEF are, but they are alternate; .. AD is parallel to EF.

PROP. 73, THEOR.

The square of the side of an equilateral triangle (DIT) inscribed in a circle is equal to three times the square of the radius,

Fig. 72.

Draw the diagonal IA, it evidently bisects the angle LIT and ... the side LT, join LA and LF.

Then the angle LFA is

LAF is to LIT ... LF is

2

to LIT, also the angle

2

=

to

to LA: but IA is to 4 FA or 4 FL2 and IA2 is = IL2 + LA2 and LA is to LF2 ... IL2 is 3 FL2.

=

=

PROP. 74, THEOR.

The right line (IA) drawn from one angle of an equilateral triangle inscribed in a circle to any point in the opposite circumference is equal to the sum of the lines (LA TA) drawn from the other two angles to the same point.

Fig. 72.

Make AF to AL, join LF.

Then since the angles LAI LTI are in the same segment, they are.. LAF is an angle of an equilateral triangle: also TAI is to ILT.. LAT is = to the sum of FLA LAF, also LFI is to them ..

2 A

+

LFI is to LAT; and the angles ILT FLA being , if FLT be taken away ILF will be to TLA, and since TLA is to ILF and LFI to LAT and the sides LI LT, FI shall be to AT.. LA and AT are together to IA.

PROP. 75. THEOR.

If perpendiculars (IL IE IF) be drawn from any point (I) in the circumference of a given circle to the sides of an inscribed triangle, the points (F,E,L) where those perpendiculars cut the sides shall be in directum.

Fig. 74.

Join ID IA LE and EF.

Then, since the angles IFD IED are to two right angles, IFDE might be inscribed in a circle, ... the angles IDF IEF are = being in the same segment; and since the angles IEA ILA are right, IELA might be inscribed in a circle.. IEL IAL are together = to two right angles; and since IDBA is inscribed in a circle, the angle IDF is to IAL,.. IEF is to IAL, .. IEF and IEL are together to two right angles, .. EF and EL are in directum.

PROP. 76, PROB.

Given of any triangle, the vertical angle and one of the sides containing it; to construct the triangle when the line drawn from the vertex to the bisection of the base, makes a given angle with the base. Fig. 75.

Let IE be to the given side, on it describe a segment of a circle capable of containing the supplement of the given angle at the base, make IEL= to the vertical angle, bisect EI in D, draw DF parallel to EL, join IF and produce it to meet EL in L, IEL is the required triangle.