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lines DA and DB will be the half of AB, and the whole of the interior arc half of the exterior.

PROP. XVI. THEOR.

The angles in the same segment of a circle are equal.

Let ADB be the segment of a circle; the angles AFB, AGB contained in it, or which stand on the same opposite portion AEB of the circumference, are equal to each other.

A

D

F

G

B

E

F D

B

For join CA, CB. The angle ACB, or its reverse at the centre, and terminated by the arc AEB, is double of the angle AFB or AGB at the circumference (III. 15.); these angles AFB, AGB, which stand on the same arc AEB, are, therefore, in every case, the halves of the same central angle ACB, and are consequently equal to each other.

E

Cor. Hence equal angles at the circumference must stand on equal arcs; for their doubles or the central angles, being equal, are terminated by equal arcs (III. 12.). Hence also equal angles that stand on the same base, have their vertices in the same segment of a circle.

Schol. Hence the ordinary construction of theatres, the seats being disposed in large arcs of a circle, so that the stage may to each spectator subtend an equal angle, or present always the same visual magnitude.

PROP. XVII. THEOR.

The opposite angles of a quadrilateral figure contained within a circle, are together equal to two right angles.

Let ABCD be a quadrilateral figure described in a circle; the angles A and C are together equal to two right angles, and so are those at B and D.

B

C

For join EB and ED. The angle BED at the centre is double of the angle BCD at the circumference (III. 15.); and for the same reason, the reverse angle BED is double of BAD. Consequently the angles BCD and BAD are the halves of angles about the point E, which make up four right angles;

E

wherefore the angles BCD and BAD are together equal to two right angles.

In the same manner, by joining EA and EC, it may be proved, that the angles ABC and ADC are together equal to two right angles.

Cor. 1. Hence it is evident from Prop. I. 16., that a circle may be described about a quadrilateral figure which has its opposite angles equal to two right angles.

Cor. 2. Hence if one side of a quadrilateral figure inscribed in a circle be produced, it will form an exterior equal to the opposite angle.

Cor. 3. Hence the angles at the base of a triangle inscribed in a circle, are together equal to an angle contained in the segment opposite to its vertex.

PROP. XVIII. THEOR.

Parallel chords intercept equal arcs of a circle.

Let the chord AB be parallel to CD; the intercepted arc AC is equal to BD.

For join AD. And because the straight lines AB and CD are parallel, the alternate angles BAD and ADC are equal (I. 22.); wherefore these angles, having their vertices in the circumference of the circle, must stand on equal arcs (III. 16. cor.),

B

D

and consequently the arcs AC and BD are equal to each other.

Cor. Hence, conversely, the straight lines which intercept equal arcs of a circle are parallel; and hence another mode of drawing a parallel through a given point to a given straight line.

PROP. XIX. THEOR.

The angle in a semicircle is a right angle, the angle in a greater segment is acute, and the angle in a smaller segment is obtuse.

Let ABD be an angle in a semicircle, or that stands on the semicircumference AED; it is a right angle.

For ABD, being an angle at the circumference, is half of the angle at the centre on the same base AED (III. 15.); it is, therefore, half of the angle ACD formed by the diverging of the opposite portions CA, CD of the diameter, or

E

B

D

or half of two right angles, and is consequently equal to one right angle.

Again, let ABD be an angle in a segment greater than a semicircle, or which stands on a less arc AED than the semicircumference; it is an acute angle.

For join CA, CD. The angle ABD is half of the central angle ACD, which is evidently less than two right angles; wherefore ABD is less than

one right angle, or it is acute.

But the angle AED, in the smaller segment, is obtuse. For AED stands on the arc ABD, which is greater than a semicircumference, and is the base of an angle at the

B

D

centre, the reverse of ACD, and greater, therefore, than two right angles; AED is hence an obtuse angle.

Cor. Hence conversely the arc which contains a right angle must be a semicircle.

Schol. From the remarkable property, that the angle in a semicircle is a right angle, may be derived an elegant method of drawing perpendiculars.

PROP. XX. THEOR.

The perpendicular at the extremity of a diameter is a tangent to the circle, and the only tangent which can be applied at that point.

Let ACB be the diameter of a circle, to which the straight line EBD is drawn at right angles from the extre

mity B; it will touch the circumference at that point.

For CB, being perpendicular, is the shortest distance of the centre C from the straight line EBD (I. 17.); wherefore every other point in this line is farther from the centre than B,

and consequently. falls without the circle.

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But EBD, drawn at right angles to the diameter, is the only straight line which can pass through the point B and not cut the circle. For were HBF such a line, the perpendicular CG let fall upon it from the centre, would be less than CB (I. 17.), and must therefore lie within the circle; consequently HBG, being extended, would again meet the circumference.

Cor. Hence a straight line drawn from the point of contact at right angles to a tangent, must be a diameter, or pass through the centre of the circle.

Scholium. The nature of a tangent to the circle is easily discovered from the consideration of limits. For suppose the straight line DE, extending

ways,

H

K

מן

B

D

both to turn about the extremity B of the diameter AB; it will cut the circle first on the one side of AB, and afterwards on the other. But the arc AH being less than a semicircumference, the angle HBA which the line D'E' makes with the diameter is acute (III. 19.); and, for the same reason, the angle KBA is acute, and consequently its adjacent angle D'BA is obtuse. Thus the revolving line DE, when it meets the semicir

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