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Join O, C the centres of the circles, OC will pass through G: let it meet AE in F. The vertically opposite angles at F being equal, and also the alternate angles OAF, FEC, the triangles AOF, FCE are equiangular, :. AO : CE :: OF : FC,

=

comp. AO CE : CE :: OF + FC : FC. But OC AO+ CE, and .. FC = CE = CG, and consequently F and G coincide, or AE intersects OC in the point G, i. e. it passes through the point of con

tact.

(36.) If two circles touch each other, and also touch a straight line; the part of the line between the points of contact, is a mean proportional between the diameters of the circle.

Let AEB, CED be two circles touching each other in E, and a straight line AC in A and C; draw the diameters AB, CD; AC is a mean proportional between AB and CD.

B

D

Join AD, BC; these lines (ii. 25.), pass through the point of contact C. And since CA touches the circle in A, from which AE is drawn, the angle CAD is equal to the angle in the alternate segment ABE; also the angle ACD being a right angle is equal to the angle CAB, .. the triangles ACD, ABC are equiangular, and

BA: AC: AC: CD.

(37.) If two circles touch each other externally, and the line joining their centres be produced to their circumferences; and from its middle point as a centre with

any radius whatever a circle be described, and any line placed in it passing through the point of contact; the parts of the line intercepted between the circumference of this circle and each of the others will be equal.

Let ABC, DCE be two circles which touch each other externally in C, and let AFE be the line joining their centres and produced to the circumferences in A and E. Bisect AE in F, and with the centre Fand any radius, let a circle GHK be de

K

K

CF

H

scribed, and in it any line GCH drawn through C meeting the circumferences of the circles in B and D; then will GB= DH.

an angle in a And since IF

Join AB, DE, and draw FI parallel to AB; it will be perpendicular to GH since ABC is semicircle, and .. GH is bisected in I. is parallel to AB,

(Eucl. vi. 2.) AF : BI :: FC : IC,

also the triangles ICF, ECD being similar,

FC: CI:: EF: ID,

.. (Eucl. v. 15.)

AF: BI :: EF : ID.

But AF FE, .. BI = ID,

=

and it has been shewn that GI=IH, whence GB=DH.

(38.) If from the point of contact of two circles which touch each other internally any number of lines be drawn; and through the points, where these intersect the circumferences, lines be drawn from any other point in each circumference, and produced to meet; the angles formed by these lines will be equal.

Let the two circles ABC, DEC touch each other internally in C, from which let any lines CA, CB be drawn; and taking any two points G and F, through E and B draw GEI, FBI, and through D and A draw GDH, FAH; if those lines

E

B

H

meet, the angle at I will be equal to the angle at H. For the angles CBF, CAF standing on the same circumference CF, are equal, .. the angle IBE is equal to HAD. Also the angles CEG, CDG, standing on the same circumference CG, are equal, and ... the angle IEB is equal to the angle HDA; .. the triangles IEB, HDA have two angles in each equal, and consequently the remaining angles are equal, i. e. EIB= DHA.

(39.) If two circles touch each other internally, and any two perpendiculars to their common diameter be produced to cut the circumference; the lines joining the points of intersection and the point of contact are proportionals.

Let the two circles ACB, AEI touch each other internally in the point A, from which let the common diameter AIB be drawn, and from any two points G, H let per

E

GHI

pendiculars GC, HD meet the circumferences in C, D,

E,

F; join AC, AD, AE, AF; these lines are proportional. For since AB : AD :: AD : AH,

AB: AH in the duplicate ratio of AB : AD.

For the same reason,

AG: AB in the duplicate ratio of AC : AB,

.. 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 contact are in continued proportion.

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 per

F

pendicular BF meet the circumferences in E and F; join AE, AF; then AB : AE :: AE : AF.

:

For (Eucl. vi. 8.) AB AF :: AF : AD. But by the hypothesis AC: AB :: AD: AC, .. AC: AF :: AF: AC,

whence AF = AC.

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.)
DA: AB :: AB : AE,

For the same reason, AB : AH :: AF : AB,
.. ex æquo DA: AH :: AF: AE,

H

but AF= AE, .. DA=AH and consequently DEHF. 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

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