Join BD, BE, BC. Since the angle CAB is in each of the two equal circles, the circumferences BD, BC on which it stands are equal, and.. the straight lines BD, BC are equal, and consequently the angle BDE is equal to the angle BCE; and the angle BED in a semicircle is a right angle, and .. equal to BEC, and BE is common to the two triangles BED, BEC, .. DE=EC.

(32.) If two circles touch each other externally or internally; any straight line drawn through the point of contact, will cut off similar segments.

D

Let the circles ADC, BCE A touch each other in the point C, and let any line ACB be drawn through the point of contact; it will cut off similar segments.

B

E

For draw the diameters CD, CE, and join AD, BE. Then DCE being a straight line (Eucl. iii. 12.), the angle ACD is equal to BCE, and DAC- CBE each being in a semicircle, and .. a right angle, whence the angles ADC, CEB are equal, and the segments ADC, CEB similar; and .. the segments AC and CB are also similar.

(33.) If two circles touch each other externally or internally; two straight lines drawn through the point of contact will intercept arcs, the chords of which are parallel.

Let the two circles ACD, ECB touch 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.

Then (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.

Gr

B

E

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. 40+ 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.

B

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

line

any radius whatever a circle be described, and any 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

G

K

K

CF

I

H

D/H

AE in F, and with the centre F and 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.

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: CL:: 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 circumferences; 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

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,