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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. 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; 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 cirele 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 e
qual to the opposite angle. Cor. 3. Hence the angles at the base of a triangle in
scribed in a circle, are together equal to an angle contain
ed 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.
and consequently the arcs AC and BD are equal to each
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, A half of the angle ACD formed by the diverging of the opposite portions CA, CD of the diameter, or
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 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 cir-
tact at right angles to a tangent, must be a diameter, or
pass through the centre of the circle. -
cumference AHB, makes an acute angle with the diameter; but when it comes to meet the opposite semicircumference, it makes an obtuse angle. In passing, therefore, through all the intermediate gradations from minority to majority, the line DE must find a certain individual position in which it is at right angles to the diameter, and cuts the circle neither on the one side nor the other.
PROP. XXI. THEOR.
If, from the point of contact, a straight line be drawn to cut the circumference, the angles which it makes with the tangent are equal to those in the alternate segments of the circle.
Let CD be a tangent, and BE a straightline drawn from the point of contact, cutting the circle into two segments BAE and BFE; the angle EBD is equal to EAB, and the angle EBC to EFB.
For draw BA perpendicular to CD (I. 5. cor.), join AE, and from any point F in the opposite arc, draw FB and FE. . . . . . Because BA is perpendicular to the tangent at B, it is a diameter (III. 20. cor.), and consequently AEFB is a semicircle ; wherefore AEB is a right angle (III. 19.), and the remaining acute angles BAE, ABE of the triangle, being together equal to another right angle, are equal to ABE and EBD, which compose the right angle ABD. Take the angle ABE away from both, and the angle BAE remains equal to EBD.