(85.) If from any point in the diameter of a semicircle, a perpendicular be drawn, meeting the circumference, and on it as a diameter a circle be described, to the centre of which a line is drawn from the farther extremity of the diameter of the semicircle, cutting its circumference; and through the point of intersection another line be drawn from the extremity of the perpendicular, meeting the diameter of the semicircle; this diameter will be divided into three segments which are in continued proportion.

From any point D in the diameter AC of the semicircle ABC, let a perpendicular DB be drawn, on which describe a circle DBG. Find its centre H, join HC cutting the circumference in G; join BG and produce it to E; AD: DE :: DE: EC.

A D

E

Join BC, and draw EF parallel to BD; join DG, GF, AB. Since EF is parallel to DB, and DB is bisected in H, .. EF is bisected in I. Also the angle HBG is equal to the alternate angle GEI, and BHG= GIE,.. the triangles BHG, GIE are equiangular,

and BH: HG :: EI: IG,

or HD: HG :: FI: IG,

i. e. the sides about the equal angles are proportional, .. the triangle FGI is equiangular to HDG, and the angle FGI is equal to HGD; whence HI being a straight line, FDG is also. Again the angle GDE is equal to the angle in the alternate segment DBG,

whence the triangles BDE, FDE are similar,

.. BD DE :: DE: EF,

but AD: DB :: EF : EC, since the tri

angles ADB, EFC are similar,

.. AD: DE :: DE: EC.

(86.) If from a point without a given circle, any two lines be drawn cutting the circle; to determine a point in the circumference, such that the sum of the perpendiculars from it upon these lines may be equal to a given line.

H

B

From the point A without the circle BDC let AB, AC be drawn cutting the circle; draw AF perpendicular to AB, and equal to the given line; FG parallel to AB, and meeting AC produced in G; from G draw GH bisecting the angle AGF, and (if the problem be possible) meeting the circle in H; H is the point required..

Through H draw KL perpendicular to AB, and HI perpendicular to AC; then the angle KGH being equal to HGI, and the angle at K to the angle at I, and the side HG, opposite to one of the equal angles in each common, HK=HI; whence HI and HL together are equal to HK and HL together, i. e. to AF, i. e. to the given line.

If GH cuts the circle, there are two points which answer the conditions.

(87.) If two circles cut each other, and any two points be taken in the circumference of one of them, through which lines are drawn from the points of intersection and produced to the circumference of the other; the straight lines joining the extremities of those which are drawn through the same point, are equal.

Let the two circles ACB, AEB. quis cut each other in A and B, and in

ACB let any two points C and D

B

be taken, through which draws ACG, BCE, ADH, BDF; and no ow join EG, FH; EG=FH.

For the angles CAD, CBD being on the same circumference CD are equal to one another, ... the circumference EF is equal to the circumference GH. Add to each FG, and the circumference EFG is equal to FGH, ... (Eucl. iii. 29.), the straight line EG=FH.

(88.) If two circles cut each other; the greatest line that can be drawn through the point of intersection is that which is parallel to the line joining their centres.

Let the two circles ABE, AFD cut each other in A. Join O, C their centres, and through A let BAD be drawn parallel to OC; BAD is greater than any other line EAF which can be drawn through A.

B

E

H

G

K

Draw OG, CH, perpendicular to BD, and OI, CK perpendicular to EF. Then AG being half of AB, and AH of AD, GH is half of BD. For the same reason IK is half of EF. Draw CL parallel to EF, and therefore at right angles to OI, and equal to IK. Then since the angle CLO is a right angle, it is greater than COL, ..the side CO is greater than CL, and GH than IK, consequently BD is greater than EF. In the same way BD may be shewn to be greater than any other line drawn through 4.

L

(89.) Having given the radii of two circles which cut each other, and the distance of their centres; to draw a straight line of given length through their point of intersection, so as to terminate in their circumferences.

F Let the two circles AFD, BGD cut each other in D; on OC, the

line joining their centres O and C,

E

G

describe a semicircle CEO; and in it, from C place CE equal to half the given line, and through D draw FDG parallel to it; FG will be the line required.

Through E draw OEH, which (Eucl. iii. 31.) will be perpendicular to FG; and draw CI parallel to OH, and.. perpendicular to DG; then (Eucl. iii. 3.) FD and DG are bisected in H and I, and .. FG is double of HI; but HECI being a parallelogram, HI = EC; .. FG is double of EC, and consequently equal to the given line.

(90.) If two circles cut each other; to draw from one of the points of intersection a straight line meeting the circles, so that the part of it intercepted between the circumferences may be equal to a given line.

Let the two circles ABC, ADB cut each other in A and B. Join AB, and draw BC touching the circle ABD. Join AC; and take AF a fourth proportional to BC, BA and the given line; join BF, and produce it to E; BFE will be the line required.

D

B

Since the angle AFB together with the angle in the segment ADB or (Eucl. iii. 32.) its equal ABC, are

equal to two right angles, i. e. to the angles AFB, AFE, :. ABC=AFE; and ACB=AEF, being in the same segment,.. the triangles ACB, AEF are equiangular, BC :: AF: FE,

and AB

but AB

BC: AF: the given line,

whence FE is equal to the given line.

(91.) If two circles cut each other; to draw from the point of intersection two lines, the parts of which intercepted between the circumferences may have a given ratio.

Let the two circles ABC, ABD cut each other in A and B; in the circle ABD place BE, BF, which have to each other the given ratio; join AE, AF, and produce AE to G; EG will have to HF the given ratio.

F

G

B

D

Draw the diameters AC, AD; join GB, BH, BC, BD; then ADBE being a quadrilateral figure inscribed in a circle, the angles AEB, ADB are equal to two right angles, and .. equal to BEA, BEG, :. BEG=BDA=BFA. And since AGBH is a quadrilateral figure inscribed in a circle, AHB, AGB are equal to two right angles, i. e. to AHB, BHF, .. AGB BHF; hence the triangles GBE, FBH are equiangular,

.. GE HF:: BE BF, i. e. in the given ratio,

:

(92.) If a semicircle be described on the common chord of two intersecting circles, and a line be drawn