(76.) If a chord and diameter of a circle intersect each other at any angle, and a perpendicular to the chord be drawn from either extremity of it, meeting the circumference and diameter produced; the whole perpendicular has to the part of it without the circle, the same ratio that the greater segment of the chord has to the less.

Let the diameter AB and chord DE intersect each other at C; and from D draw DG perpendicular to DE,

meeting AB produced in G;

H

then DG GF :: DC :: CE.

:

Through Fdraw FI parallel to DE, and meeting the diameter in I. Join FE, cutting the diameter in O. Since the angle FDE is a right angle, FE is a diameter and O is the centre. And since the angle IFO is equal to the alternate angle OEC, and the angles at O ́are equal, and FO=OE, .. the triangles OFI, OEC are equal, and CE=FI. And since FI is parallel to DC,

(Eucl. vi. 2.) GD : GF :: DC : (FI=) CE.

In a similar manner it may be shewn that gH : gE:: DC: CE.

(77.) If from the extremities of any chord of a circle, perpendiculars to it be draum and produced to cut a diameter; and from the points of intersection with the diameter lines be drawn to a point in the chord, so as to

make equal angles with it; these lines together will be equal to the diameter of the circle.

Let AB be any chord of the circle ABC; draw AE and BF perpendicular to it, meeting the diameter CD in E and F ; from which let the lines EH, FH, be drawn

making equal angles with AB; EH and HF together are equal to CD.

Take O the centre, and join BO, and produce it; it will meet AE produced in G. Produce EH, FB to meet in I. Then since the angle IHB=AHE=FHB, and HB is perpendicular to FI, the triangles FHB, HBI are equal, and FH=HI. And since EG is parallel to FB, the angle EGO OBF, and the vertical angles at O are equal, and GO=OB, .. EG=FB=B1; whence EI= GB, and.. EH, HF together are equal to EI i. e. to GB or CD the diameter of the circle.

(78.) If from a point without a circle two straight lines be drawn, one of which touches and the other cuts the circle; a line drawn from the same point in any direction, equal to the tangent, will be parallel to the chord of the arc intercepted by two lines drawn from its other extremity to the former intersections of the circle.

From the point A let AB, AD be drawn, of which AB touches the circle BCD, and AD cuts it, and draw AE. B AB, in any direction; join CE, DE, cutting the circle in F and G; the chord FG will be parallel to AE.

K

D

G

Because (Eucl. iii. 36.) DA: AB :: AB: AC, and AE= AB, .. DA: AE :: AE: AC, i. e. the sides about the angle A of the triangles ADE, ACE are proportional, .. (Eucl. vi. 6.) the triangles are equiangular, and the angle AEC is equal to the angle ADE. But since CDGF is a quadrilateral figure in the circle, the angles CDG, CFG are equal to two right angles, i. e. to EFG, CFG, .. CDG = EFG, whence AEF=EFG, and FG is therefore parallel to AE.

(79.) If from a point without a circle, two straight lines be drawn touching it, and from one point of contact a perpendicular be drawn to that diameter which passes through the other; this perpendicular will be bisected by the line joining the point without the circle and the other extremity of the diameter.

Let DA, DB be drawn from a point D without the circle ABC, touching it in A and B, and from B let BE be drawn perpendicular to AC the diameter passing through A; join CD; BE is bisected by CD in the point F.

G

D

F

E

A

Then

For produce AD and CB to G; join AB. since DA=DB, the angle DAB is equal to the angle DBA. Now the angle ABG, being a right angle, is equal to BAG, BGA, of which ABD=BAG, :. DBG DGB, and DG= DB=DA; and since AG is parallel to EB,

BF: GD :: CF : CD :: EF : AD,

and GD=DA, .. BF=FE.

(80.) If any chord in a circle be bisected by another, and produced to meet the tangents drawn from the extremities of the bisecting line; the parts intercepted between the tangents and the circumferences are equal.

Let AB be bisected in E by CD; and to C and D let tangents be drawn, meeting AB produced in F and G; AF is equal to BG.

Find

the centre; join OC,

OD,

E

D

OE, OF, OG. Since OE is drawn from the centre to the point of bisection of AB (Eucl. iii. 3.). the angle OEF is a right angle; and the angle OCF is a right angle (Eucl. iii. 18.); .. a circle may be described about OEFC. Also since ODG and OEG are right angles a circle may be described about OEDG; and the angle DOG is equal to the angle DEG in the same segment; but DEG is equal to FEC, i. e. to FOC, .. DOG = FOC; and ODG, OCF are equal being right angles, and OC=OD, .. OF=OG, and consequently FE= EG. But AE=EB, .. FA = BG.

(81.) If one chord in a circle bisect another, and tangents drawn from the extremities of each be produced to meet; the line joining their points of intersection will be parallel to the bisected chord.

Let AB be bisected by the line CD in E, and let the tangents AF, BF meet each other in F, and DG, CG in G. Join GF; GF is parallel to AB.

Join AO, CO, GO, FO; then

D

B

GO bisects CD in H, and OHE is a right angle; for the same reason FO passes through E, and AEO is a right angle. And since FAO is a right angle (Eucl. iii. 18.), and from A, AE is drawn perpendicular to the base,

(Eucl. vi. 8. Cor.) FO: OA:: OA: OE, for the same reason,

(OC=) OA: ÓG :: OH : (OC=)0A, .. ex æquo per. FO: OG: OH: OE,

.. the sides of the triangles FOG, OHE about the common angle are proportional, and consequently the triangles are equiangular, and the angle GFO equal to EHO, and .. a right angle, and equal to the alternate angle FEB,.. AB is parallel to GF.

(82.) If from a point without a circle two lines be drawn touching the circle, and from the extremities of any diameter lines be drawn to the points of contact, cutting each other within the circle; the line produced, which joins their intersection and the point without the circle, will be perpendicular to the diameter.

D

O F

B

From the point P without the circle ABC let there be drawn two tangents PC, PD. From A and B the extremities of a diameter, draw AD, BC to the points of contact, intersecting each other in E; join PE, and produce it to F; PF is perpendicular to AB.

Take O the centre; join CO, DO, CD, DB. Since CPD, COD, are together equal to two right angles,