CPD is equal to AOC, BOD together, i. e. to twice ADC, BCD together, ... CPD is equal to the angle at the centre of a circle passing through C, E, and D; and since PC-PD, P is the centre itself; .. PE=PD, and the angle PED is equal to the angle PDE. But the angle DBA=PDE=PED=AEF, and the angle at A is common, .. AFE=ADB, and (Eucl. iii. 31.) is ... a right angle.

(83.) If on opposite sides of the same extremity of the diameter of a circle equal arcs be taken, and from the extremities of these arcs lines be drawn to any point in the circumference, one of which cuts the diameter, and the other the diameter produced; the distances of the points of intersection from the extremities of the diameter are proportional to each other.

On opposite sides of the point A in AB the diameter of the circle ABC let equal arcs AC, AD be taken; from C and D let CE, DE be drawn to any point E in the circumference, of which CE cuts AB produced in F, and DE cuts AB in G; then will AF: FB :: AG : GB.

Join AE, BE, and through B draw HBI parallel to AE. Since AEB is a right

..

H

angle, CEA and BEI are together equal to a right angle, and equal to AED, DEB; and since AC AD, CEA = AED, .. BEI= BED. Again since AE is parallel to IH, the angle EIB is equal to CEA = AED= the alternate angle EHB, .. the two triangles EIB, EHB having two angles in each equal, and one side EB

common are equal, and BI=BH. And from the similar

triangles AGE, BGH,

AG: AE: BG: (BH=) BI,

alt. AG: GB :: AE : BI :: AF: BF, since the triangles AEF, FIB are similar.

(84.) If from the extremities of any chord in a circle, perpendiculars be drawn to a diameter, and from either extremity of that diameter a perpendicular be drawn to the chord; it will divide it into segments, which are respectively mean proportionals between the segments of the diameter made by the perpendiculars.

Let AB be any chord, and CD a diameter of the circle ABC; AE, BF perpendiculars from A and B to the diameter, and DG perpendicular from D to AB; the segment GB is a mean proportional between DF and CE; and AG a mean proportional between DE and CF.

Join AC, AD, BC, BD. Then the angle DBA being equal to DCA and the angles DGB, DAC, AEC being right angles, the triangles DGB, DAC, ACE are similar; also the triangles DBF, DBC are similar; whence

DF: DB::DB: DC :: BG: AC but DB: BG:: AC: CE,

.. ex æquo DF: BG :: BG: CE.

And in the same manner it may be proved that
DE: AG :: AG: CE

(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.

B

H

F

E

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

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 1. 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, FGD 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.

AL

D

H

C

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 whick are drawn through the same point, are equal.

Let the two circles ACB, AEB cut each other in A and B, and in ACB let any two points C and D be taken, through which draw ACG, BCE, ADH, BDF; and join EG, FH; EG=FH.

G

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 B 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.

E

G

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 A.

L