.. AG AH in the duplicate ratio of AC: AD. In the same manner it may be shewn that AG AH in the duplicate ratio of AE: AF, .. (Eucl. v. 15.) the duplicate ratio of AC : AD, is the same with the duplicate ratio of AE : AF, and. AC: AD :: AE: AF (40.) If three circles, whose diameters are in continued proportion touch each other internally, and from the extremity of the least diameter passing through the point of contact, a perpendicular be drawn, meeting the circumferences of the other two circles; this diameter and the lines joining the points of intersection and con, tact are in continued proportion. F E Let AB, AC, AD the diameters of three circles touching each other in A, be in continued proportion, viz. AB : AC:: AC: AD, and from B the perpendicular BF meet the circumferences in E and F; join AE, AF; then AB For (Eucl. vi. 8.) : A AE :: AE: AF. AB AF :: AF: AD. But by the hypothesis AC: AB :: AD : AC, .. AC: AF:: AF: AC, whence AF = AC. B C And (Eucl. vi. 8.) AB : AE :: AE : AC, .. AB AE :: AE: AF. (41.) If a common tangent be drawn to any number of circles which touch each other internally; and from any point in this tangent as a centre, a circle be described cutting the others, and from this centre lines be drawn through the intersections of the circles respectively; the segments of them within each circle will be equal. Let the circles touch each other in the point B, to which let a tangent BA be drawn, and from any point A in it as a centre with any radius, let a circle EFG be described. Draw the lines AED, AFH, AGI; then will the parts DE, HF, IG be equal. B For since AB touches the circle, (Eucl. iii. 36.) For the same reason, AB: AH :: AF: AB, .. ex æquo DA: AH :: AF: AE, but AF= AE, .. DA=AH and consequently DE=HF. In the same manner it may be proved, that IG=HF or DE (42.) If from any point in the diameter of a circle produced, a tangent be drawn; a perpendicular from the point of contact to the diameter will divide it into segments which have the same ratio that the distances of the point without the circle from each extremity of the diameter, have to each other. From any point C in the diameter BA produced, let a tangent CD be drawn, and from D, draw DE perpendicular to AB; AE: EB:: AC: CB. G A E GEOMETRICAL PROBLEMS. [Sect. 2. Take O the centre of the circle, join DO; then (Eucl. iii. 18.) the angle CDO is a right angle, and .. (Eucl. vi. 8.) CO: OD :: OD : OE, or CO OA :: OA : OE, .. div. and comp. AC: CB :: AE : EB. COR. The converse may easily be proved. (43.) If from the extremity of the diameter of a given semicircle a straight line be drawn in it, equal to the radius, and from the centre a perpendicular let fall upon it and produced to the circumference; it will be a mean proportional between the lines drawn from the point of intersection with the circumference to the extremities of the diameter. From B the extremity of the diameter AB let a line BC be drawn, equal to the radius BO; and on it let fall a perpendicular OD meeting the circumference D B A in D; join DB, DA; DO is a mean proportional between DA and DB. Join DC. Then the angles BAD, BCD on the same base are equal. Also since OD bisects BC, it bisects the arc BDC, .. also the straight line BD=DC and the angle DBC=DCB, but ODA=OAD, .. the triangles ODA, DBC are similar, .. AD: DO :: (BC=) DO : DB. (44.) If from the extremity of the diameter of a circle, two lines be drawn, one of which cuts a perpen dicular to the diameter, and the other is drawn to the point where the perpendicular meets the circumference; the latter of these lines is a mean proportional between the cutting line, and that part of it which is intercepted between the perpendicular and the extremity of the diameter. Let CE be at right angles to the diameter AB of the circle ABC, and from A let AD, AC be drawn, of which AD cuts CE in F, then will AD: AC :: AC : AF. D f B G E For since the circumference AE is equal to the circumference AC, (Eucl. iii. 27.) the angle ECA is equal to the angle ADC, and the angle at A is common to the two triangles ADC, ACF, .. the triangles are similar, and AD: AC :: AC : AF. But if the point of intersection f be without the circle, draw dH parallel to CG, then, as before, the angle HdA is equal to ACd, and the angle at A common to the triangles AHd, ACd, .. Ad: AC :: AH : Ad :: AC : Aƒ. (45.) In the diameter of a circle produced, to determine a point, from which a tangent drawn to the circumference shall be equal to the diameter. From A the extremity of the diameter AB, draw AD at right angles and equal to AB. Find the centre O, join OD cutting the circle in C; and through C draw CE at right angles to OD meeting BA produced in E. B GEOMETRICAL PROBLEMS. [Sect. 2. Then because the angle OAD is equal to OCE, each being a right angle, and the angle at O is common to the two triangles OAD, OCE, and OA=OC, .. AD= CE. But AD was made equal to AB, .. CE=AB, and E is the point required. (46.) To determine a point in the perpendicular at the extremity of the diameter of a semicircle, from which if a line be drawn to the other extremity of the diameter, the part without the circle may be equal to a given straight line. From B the extremity of the diameter of the semicircle ADB, let a perpendicular BC be drawn; in which take BE equal to the given line; and on it as a diameter describe a circle; through the centre of which draw AGF, and with A as centre and radius AF describe a circle cutting BC in C. Join AC; CD is equal to the given line. Join BD. Then BD being perpendicular to AC, (Eucl. vi. 8. Cor.) and (Eucl. iii. 36.) .. ex æquo, AB: AF :: AG: AB, AC: AF :: AG: AD, whence AG=AD, and .. DC=GF=BE. (47.) Through a given point without a given circle, to draw a straight line to cut the circle, so that the two perpendiculars drawn from the points of intersection to that diameter which passes through the given point, may |