Let the two circles ACD, ECB touch F each other in C, and let ABC, DEC be any two lines drawn through the point of contact. Draw the tangent FCG; and join AD, EB; AD, EB are parallel.

G

For (Eucl. iii. 32.) the angle ADC = (FCA =) BEC, whence (Eucl. i. 28.) AD is parallel to BE.

(34.) If two circles touch each other internally or externally; any two straight lines drawn through the point of contact and terminated both ways by the circumferences will be cut proportionally by the circumference.

Let the two circles touch each other in C, (see last Fig.) and let ACB, DCE be any two lines drawn through the point of contact; then it may be shewn (as in the last prop.), that AD is parallel to BE, and the triangles ACD, BCE are similar,

.. AC; CB :: DC : CE,

(35.) If two circles touch each other externally, and parallel diameters be drawn; the straight line joining the extremities of these diameters, will pass through the point of contact.

Let ABG, DGE be two circles touching each other externally in the point G; and let AB, DE be parallel diameters; join AE; AE will pass through G.

B

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 circles.

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.

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.

K

K

CF

IH

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

D/H

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

EX

meet, the angle at I will be equal to the angle at H.

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 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 circumferences; the lines joining the points of intersection and the point of contact are proportionals.

E

GHI

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 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,

per

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 perpendicular BF meet the circumferences in E and F;

:

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