GEOMETRY. 
at FGHIK the duplicate ratio of AB 
‘to '. 
219 
angle DEF also three times ; and, in general, what- Problems to 
ever number of times the arc AB contains some part of Sect. IV. 
Pror. XXVIII. Tueor. 
The segments of two chords AB, CD, which cut each 
other within a circle, are reciprocally proportionals, that 
iss AO: DO:: CO: OB. 
Join AC and BD: In the triangles AOC, BOD, the 
vertical angles at O are equal; also the angle A= D, 
‘and C= B (17. 2.); ‘ore the triangles are similar, 
and the ho us sides proportionals, that is, AO : 
OD::CO: OB. 
Cor. Hence the rectangle AO x OB is equal to the 
rectangle CO x OD (22.) That is, the rectangle con- 
tained by the segments of the one chord is equal to the 
rectangle contained by the segments of the other. 
Pror. XXIX. Tueor. 
If two chords BA, CD in a circle be produced to in- 
tersect each other without it ; the distances of the ex- 
tremities of the chords from’ their mutual intersection 
are pn pepe! proportionals, that iss AO: DO:: 
CO BY. 7 
For, joining BD and AC, the triangles OAC, OBD 
have the angle O common, and besides, the angle B=C 
(17. 2.), therefore the triangles ate similar, and the ho- 
orate oo a at that is, AO: OD:: CO: 
B 
Cor. Hence the rectangle AO x OB is equal to 
the rectangle CO x OD (22.) 
Prop. XXX. Tueor. 
If from any point O in the prolongation of a chord 
CD, a tangent OA be drawn to the circumference, the 
tangent is a mean proportional between the distances 
of the intersection from the extremities of the chord. 
That is, CO: OA :: OA: OD. 
__ For if DA and AC be joined, the triangles OAD and 
OAC have the angle at O common, besides the angle 
OAD=C (20. 2.) ; therefore the two triangles are simi- 
lar, and hence CO: OA:: OA: OD. 
Cor. Hence we have AO?=CO x OD, (Cor. 22.) 
Scuotium. The three preceding propositions have a 
great affinity, In fact, a constitute an individual 
property of the circle; for when two of the intersec- 
tions unite, the chord becomes a tangent. 
Prop. XXXI. Tueor. 
ACB, DEF at the centres are to.each other as the arcs 
AB, DF of the circles, intercepted between the lines 
which contain the angles. 
Let us suppose that the arc AB contains three such 
equal parts as DF contains five: Let Ap, pq, qB be 
the equal parts in AB, and Dr, 7's, &c. the equal parts 
in DF ; draw the lines Cp, Cq, Er, Es, &c.; the 
angles AC p, pCq, gCB, DEr, &c. are all equal (15.2:) 
therefore as the arc AB-contains jth of the are DF three 
‘times, the angle ACB will evidently contain }th of the 
In the’same circle, or in equal circles, any angles. 
the arc DF, the same number of times will the angle 
ACB contain a like part of the angle DEF. 
PROBLEMS RELATIVE TO SECT, IV. 
Prosiem I. 
To divide a given straight line into any ‘number of 
equal parts, or into parts proportional to given lines, 
1. Let it be og te to divide the line AB into five Fig. 118. 
equal parts. Through the extremity A draw an inde- 
finite straight line AG ; and in this line, take five equal 
distarices AC,CD,DE, EF, andFG, of any length. Join 
BG, and draw CI parallel to GB, then AI will be the 
fifth part of AB, and the distance AI being set off five 
times from A, the line AB will be divided into five equal 
parts at the points I, K, L, M, as required. For the 
sides AG, AB are: cut proportionally in C and I, 
(15. 4.) ; and as AC is one-fifth of AG, AI will also 
be one-fifth of AB. 
Next let it be . o. 
parts proportional to the lines P, Q, R, (Fig. 119. 
From The sdicandy A, draw the Salohisin pecaaghe 
line AE, and take AC=P, CD=Q, DE=R. Join 
the extremities E and B, and draw CI, DK, parallel to 
EB ; then the line AB shall be divided into parts AI, 
IK, KB proportional to the given lines P, Q, R. 
For because of the parallels AC: CD:: AI: 1K (15.4.), 
and by composition, AD : DC:: AK: KI; again, 
DE:AD::KB: KA ; therefore, ex equo, DE: DC:: 
KB:KI, and so on; since then it appears that AC: Al:: 
CD: 1IK:: DE: KB, the of the line AB have 
to.each other the ratios of the lines AC, CD, DE, that. 
is of P, Q, and R. 
Pros, Ii. 
To find a fourth proportional to three given lines pig. 190. 
A, B,C. 
Draw the two indefinite lines DE, DF, making any 
angle. On DE take DA=A, and DB=B; and 
on DF take DC=C; join AC, and through B 
draw BX parallel to AC; then shall DX be the fourth 
proportional required. For since BX is parallel to 
AC, DA: DB::DC:DX (15. 4.); therefore DX is 
the fourth proportional required. 
Cor. By this problem, a third proportional to two 
given lines may be found ; for it will be the same as a 
fourth proportional to A, B and B.. 
Pros. III. 
_ To find a mean proportional: between two.given lines Fig 121. 
A and B. 
On the indefinite line DF take DE=A, and EF—B: 
on DF as a diameter describe a semicircle DGF; at the 
point E draw EG perpendicular. to the diameter, meet- 
ing the circumference in G; then shall EG be the 
mean proportional required. Join GD and GF; the 
triangle DGF is right angled at G, for G is an angle 
in.a semicircle (19. 2.); therefore the perpendicular GE 
to divide the line AB into: Fig. 119. 
