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Let ABCD be a four-sided plane figure, having its angular points A, B, C, D, in the circumference of a circle; then ABC+4 ADC=two right angles; and likewise BAD+ ▲ BCD = two right angles.

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triangle ABC, ABC+ BAC + ACB= two right angles (37); but by (52) ≤ BAC = 4BDC, being angles in the same segment' BADC. Also ACB = LADB, being angles in the same segment' ADCB; .. 4ABC + < BDC + ‹ ADB = two right angles; and ▲ BDC + 2 ADB = LADC (22), .. 4 ABC+4 ADC = two right angles.

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In the same manner it may be shewn that BAD+ < BCD = two right angles.

COR. Hence no parallelogram except a rectangle can be 'inscribed' in a circle; because the opposite angles in every parallelogram are equal to one another; and therefore in this case each of them must be a right angle.

54. PROF. VI. The angle in a segment' equal to a semi-circle is a right angle; in a segment greater than a semi-circle is less than a right angle; and in a segment less than a semi-circle is greater than a right angle.

Let ABCD be a circle of which AB is a diameter, and the centre; draw the

chord BD dividing the circle into two segments, viz. BAD greater than, and BCD less than, a semi-circle; join AD, DC, CB. Then L ADB 'in a semi-circle' is a right angle; L BAD 'in a segment' greater than a semi-circle is less than a right angle, and BCD in

B

a segment' less than a semi-circle is greater than a right angle.

Join OD; then since OA = OD, ¿OAD= ¿ODA (26); and since OB OD, ▲ OBD=2 ODB, .:. ▲ ADB = ‹ÒAD + < OBD; add to these equals ADB, then twice ‹ ADB

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= ¿OAD + ¿OBD + ‹ ADB ; .. twice 4 ADB = two right angles, (since ‹ OAD, ▲ OBD, and ▲ ADB are the three angles of the triangle ABD), and the halves of equal things must be equal, .. ADB = one right angle.

Also, since OAD + OBD = a right angle, ..OAD, or BAD, which is the same thing, is less than a right angle.

Again, by (53), <BCD+¿BAD = two right angles, and since BAD is less than a right angle, .. ≤BCD must be greater than a right angle.

55. PROP. VII. A straight line, drawn at right angles to a diameter of a circle from either of its extremities, lies wholly without the circumference of the circle except at that point only.

[DEF. A straight line, which lies wholly without the circumference of a circle except at one point only, is said to touch the circle, or to be a tangent to the circle, at that point.]

D

B

Let a straight line AB be drawn at right angles to AC, a diameter of the circle ACD, from the point A one of its extremities; and let O be the centre of the circle. Take E any other point in AB, distinct from A, and join OE; and let OE, or OE produced meet the circumference in F. Then AOE is a triangle; and since the three angles of every triangle are equal to two right angles, and one of them in this case, viz. OAE is a right angle, .. each of the other two angles, OEA, ‹ÃOE, is less than a right angle; that is, OAE is greater than LOEA. But the greater side is opposite to the greater angle (33); .. OE is greater than OA, the radius of the circle, that is, OE is greater than OF, or E is without the circumference. And E is any point whatever in AB except A; ... AB lies wholly without the circumference except at the point A.

L

COR. 1. The converse of this is also true, viz. that, if a straight line touch' the circle at any point, it will be at right angles to the diameter or radius through that

point. For, if possible, AB being a tangent at A, that is, every point in it except A being without the circle, suppose OAB not a right angle. From O draw OE at right angles to AB, meeting the circumference in F; then OAE is a triangle, of which OEA is a right angle, and .. < OAE less than a right angle; .. the side OA is greater than the side OE (33); but OA = OF, .. OF is greater than OE, a part greater than the whole, which is impossible. Hence the supposition that ≤ OAB is not a right angle cannot hold; .. AB must be at right angles to AO.

COR. 2. Hence, to draw a tangent to a given circle through a given point A in its circumference, find O the centre of the circle, join OA, and draw AB at right angles to 04 from the point A ; AB is the tangent required.

COR. 3. Hence, also, if a straight line touches a circle, and from the point of contact another straight line be drawn, at right angles to the former, through the circle, the centre of the circle will be in this latter line.

56. PROP. VIII. To draw a tangent to a given circle from a given point without it.

Let A be the given point from which it is required to draw a straight line touching the given circle BCD.

Find O the centre of the circle (50), and join OA; bisect OA in the point E; with centre E and radius EA describe the semi-circle AFO meeting the circle BCD in F; and join AF. AF is the tangent required.

For, joining OF, since

AFO is a semi-circle, 40FA

E

is a right angle (54),.. AF is at right angles to a radius or diameter of the circle from one of its extremities, .. AF touches the circle at the point F (55).

A second tangent may also be drawn from A by

describing the semi-circle on the other side of OA meeting the given circle in C, and then joining AC.

57. PROP. IX. If the radius of one circle be equal to the radius of another, the circles shall be equal in all respects.

For, if one of the circles be applied to', or laid upon, the other so that their centres coincide, since the radii are equal, it is plain that the circumferences will coincide throughout; and, since the circumferences coincide in every part, it is evident that they enclose the same area, that is, the circles are equal to one another.

58. PROP. X. In the same circle, or in equal circles, equal arcs have equal chords; and conversely, equal chords have equal arcs.

Let AB, CD be equal arcs of two equal circles; draw the chords AB, CD; then chord AB = chord CD. For,

B

E

D

о

find the centres E, F of the circles, and suppose the one circle to be laid upon the other, so that the centre E shall be upon F; then since the radii are equal, the whole circumference of one will coincide with the whole circumference of the other; and, without altering this coincidence, if one of them be turned round the centre, in its own plane, until the point A coincides with the point C, the point B will coincide with D, because the arc AB arc CD. So then, since the point A falls upon C, and B upon D, the straight line joining A and B must coincide with that joining C and D; that is, chord AB = chord CD.

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And what is proved of equal circles will obviously hold true for equal arcs of the same circle.

And the converse also evidently follows, viz. that equal chords in the same circle, or in equal circles, subtend equal arcs.

59. PROP. XI.

In the same circle, or in equal circles, equal arcs subtend equal angles at the centre.

Let AB, CD be equal arcs of two equal circles, whose centres are E and F. Join AE, BE, CF, DF; then L AEB = = < CFD.

B

For, joining the chords AB, CD, by (58) chord AB= chord CD; also AE= CF, and BE=DF'; .. the two triangles AEB, CFD, have all the sides of the one equal to all the sides of the other, each to each, and .. the triangles are equal in all respects (23 Cor.). Consequently AEB CDF, being the angles opposite to the equal sides AB, CD.

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60. PROP. XII. To bisect a given arc of a circle. Let ABC be the given arc. It is required to divide it into two parts in the point

E, so that the arc_AB=arc

BC.

Join AC; bisect AC in D;

from D draw DB at right A

D

angles to AC, intersecting the given arc in B. Then ABC is bisected in B.

For, drawing the chords AB, BC, the two sides AD, DB, in the triangle ADB, are equal to the two sides CD, DB, in the triangle CDB; also ADB = < CDB, since each of them is a right angle, ... the third side AB CB. But equal chords subtend equal arcs, (58), .. arc AB = arc BĊ.

L

COR. If BD be produced, it will pass through the centre of the circle. And, conversely, the line drawn from the centre, bisecting the chord, will also bisect the arc.

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