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from A through B draw ABH meeting the last chord GD in H; GH is equal to EB, CF, DG, &c. together.

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Join DF, EC, and produce EC to I. Since AE = BC, ABH is parallel to ECI. And since the arcs are equal, the lines BE, CF, DG are parallel, whence BI is a parallelogram and BE=HI. In the same manner it be shewn, that CF=ID, and so on, whatever be the number of equal arcs; hence GH is equal to the sum of BE, CF, DG.

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(28.) If the circumference of a semicircle be divided into an odd number of equal parts, and through the points which are equally distant from the diameter, lines be drawn; the segments of these lines intercepted between radii drawn to the extremities of the most remote, will together be equal to a radius of the circle..

Let the circumference of the semicircle ADB be divided into any odd number of equal parts, e. g. five, (the demonstration being the same for any odd number) in the points C, D, E, F. Join DE, CF, which are parallel, since

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they intercept equal arcs. Join OD, OE; DE and LM together are equal to the radius of the circle.

For complete the circle, and divide the opposite semicircle in the same manner; join AC, DG, EH which will be parallel to one another; CH will also be parallel to DI Hence DE OK, and OK is also equal to each of the two PM, CL, :. PM= CL, whence LM=CP which is equal to AK, since CF is parallel to AB,

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and DE + LM = KO + AK = AO_the_radius of the circle.

(29.) If from the extremities and the point of bisection of any arc of a circle, lines be drawn to any point in the opposite circumference; the sum of those drawn from the extremities will have to that from the point of bisection, the same ratio that the line joining the extremities, has to that joining one of them and the point of bisection.

Let the arc AB be bisected in C, and AB, AC joined; to any point D in the circumference draw AD, BD, CD; then AD + DB : DC :: AB : AC.

D

Draw AE parallel to CD, and let it meet BD produced in E. The angle EAD ADC = CAB (Eucl. iii. 27.); to each of these add DAB, and EAB = CAD; also ABD = ACD, .. the triangles EAB, ACD are equiangular, whence BE: CD :: BA: AC. But since CDB is equal to each of the angles AED, DAE, they are equal to one another, .. DA DE,

and AD + DB : DC :: AB : AC.

F

(30.) If two equal circles cut each other, and from either point of intersection a circle be described cutting them; the points where this circle cuts them and the other point of intersection of the equal circles are in the same straight line.

Let the two equal circles cut each other in A and B, and with the centre A and any distance AC, describe a circle FCD cutting their circumferences in C and D; C, D,B will be in a straight line. Join CB, and let it meet the circumference ADB in E. Join AE, AC. Since the angle ABC is an angle

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in each of the two equal circles, the circumference AC is equal to the circumference AE (Eucl. iii. 26.), .. the line AC is equal to the line AE; and .. E is a point in the circle FDC, and being by construction in the circumference ADB, it must coincide with D; .. CB passes through D, or C, D, B are in a straight line.

(31.) If two equal circles cut each other, and from either point of intersection a line be drawn meeting the circumferences; the part of it intercepted between the circumferences will be bisected by the circle whose diameter is the common chord of the equal circles.

Let the two equal circles ADB,

ACB cut each other in A and B ; join AB, and on it as a diameter let a circle AEB be described, and from A draw any line ADC meet

C

ing the circumferences in D and C; DC is bisected in E.

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.

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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.

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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.

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BA

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

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(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 circamferences 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.

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