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And, because Fc is equal to FD, AC to AD, and AF common to each of the triangles AFC, AFD, the angle ACF will also be equal to the angle ADF (1.21.)
But, fince AD touches the circle, and DF is a line drawn from the centre to the point of contact, the angle AdF is a right angle (III. 12.)
The angle ACF, therefore, is also a right angle; and CF produced is a diameter of the circle.
And since a right line, drawn from the end of the diameter, at right angles to it, touches the circle (III. 10.), AC will be a tangent to the circle CBD, as was to be thewn.
1. One rectilineal figure is said to be inscribed in anoa ther, when all the angles of the one are in the sides of the other.
2. One rectilineal figure is said to be described about another, when all the sides of the one pass through the angular points of the other.
3. A rectilineal figure is said to be inscribed in a circle, when all its angular points are in the circumference of the circle.
4. A rectilineal figure is said to be described about a circle, when each side of it touches the circumference of the circle.
5. A circle is said to be inscribed in a rectilineal figure, when its circumference touches every side of that figure.
6. A circle is said to be described about a rectilineal figure, when its circumference passes through all the angular points of that figure.
7. A right line is said to be placed, or applied, in a ciscle, when the extremities of it are in the circumference of the circle.
8. All plane figures contained under more than four fides are called polygons; and if the angles, as well as hdes, are all equal, they are called regular polygons.
9. Polygons of five fides, are called pentagons; those of six fides hexagons; those of seven heptagons; and
PRO P. I. PROBLEM.
To place a right line in a given circle, equal to a given right line, not greater than the diameter of the circle.
Let ABC be a given circle, and D a given right line, not greater than the diameter; it is required to place a line in the circle ABC that shall be equal to D.
Find E, the centre of the circle (III. 1.), and draw any diameter AB; then if AB be equal to D the thing required is done.
But if not, make AE equal to D (I. 3.); and from the point A, at the distance AE, describe the circle FEC, cutting the former in c.
Join the points A, C; and AC will be equal to D, as was required.
For fince A is the centre of the circle ABC, Ac is equal
But D is also equal to AE, by construction; whence Ac is, likewise, equal to D.
In the circle ABC, therefore, a right line has been placed equal to D, which was to be done.
PRO P. II. PROBLEM.
To inscribe a triangle in a given circle, that shall be equiangular to a given triangle.
Let ABC be the given circle, and DEF the given trian. gle; it is required to inscribe a triangle in the circle ABC, that shall be equiangular to the triangle DEF.
Draw the right line gh to touch the circle ABC in the point c.(III. 10.); and, make the angle HCB equal to the angle D (I. 20.), and the angle GCA to the angle E; and join AB; then will ace be the triangle required.
For, since the right line gh is a tangent to the circle, and cB is a chord drawn from the point of contad, the angle HCB will be equal to the angle cas in the alternate segment (III. 24.)
But the angle HCB is equal to the angle D, by construction ; therefore the angle CAB is also equal to the angle D.
And, in the same manner, it may be proved, that the angle CBA is equal to the angle e.
Bụt, since the angle CAB is equal to the angle D, and the angle CBA to the angle E, the remaining angle ACB will also be equal to the remaining angle F (Cor. I. 28.), and consequently the triangle AcB is equiangular to the triangle def.
Q. E. D.