5. A circle is faid to be inscribed in a rectilineal figure, when its circumference touches every fide of that figure.

6. A circle is faid to be defcribed about a rectilineal figure, when its circumference paffes through all the angular points of that figure.

7. A right line is faid to be placed, or applied, in a circle, when the extremities of it are in the circumference of the circle.

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8. All plane figures contained under more than four fides are called polygons; and if the angles, as well as fides, are all equal, they are called regular polygons.

9. Polygons of five fides, are called pentagons; those of fix fides hexagons; those of seven heptagons; and

fo on.

PROP. I. PROBLEM.

To place a right line in a given circle, equal to a given 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 fhall be equal to D.

Find 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, defcribe 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 CEF, AC is equal

to AE.

But D is alfo 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.

PROP. II. PROBLEM.

To infcribe a triangle in a given circle, that shall be equiangular to a given triangle.

G

H

Let ABC be the given circle, and DEF the given triangle; it is required to inscribe a triangle in the circle ABC, that fhall 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 ACB be the triangle required.

For, fince the right line GH is a tangent to the circle, and CB is a chord drawn from the point of contact, the angle HCB will be equal to the angle CAB in the alternate fegment (III. 24.)

But the angle HCB is equal to the angle D, by conftruction; therefore the angle CAB is alfo equal to the angle D.

And, in the fame manner, it may be proved, that the angle CBA is equal to the angle E.

But, fince 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 confequently the triangle ACB is equianglar to the triangle DEF.

Q. E. D.

PROP. III. PROBLEM.

To circumfcribe a triangle about a given circle, that fhall be equiangular to a given triangle.

Let ABC be the given circle, and-DEF the given triangle; it is required to circumfcribe a triangle about the circle ABC that fhall be equiangular to the triangle DEF.

Produce the line EF to G and H; and, at the centre I, make the angles AIB, BIC equal to the angles DEG, DFH (I. 20.); and draw the lines MK, KL, LM, to touch the circle in the points A, B, C (III. 10.); and join Ab.

Then, fince the angles IAK, KBI, are, each of them, a right angle (III. 12.), the angles BAK, KBA, taken together, will be less than two right angles.

But when a right line interfects two other right lines, and makes the two interior angles, on the fame fide, together lefs than two right angles, thofe lines will, if produced, meet each other (I. 25..Cor.)

The line MK, therefore, meets the line KL; and, if A, C, C, B be joined, the fame may be proved of the lines KL, LM and MK; confequently the figure KLM is a triangle.

And, because the four angles of the quadrilateral AIBK are equal to four right angles (Cor. I. 28.), and the angles

IAK, KBI are each a right angle, the remaining angles AIB, BKA will be equal to two right angles.

But the angles DEG, DEF are alfo equal to two right angles (I. 13.); therefore, fince the angle DEG is equal to the angle AIB (by Conft.), the remaining angle BKA will be equal to the remaining angle DEF.

And, in the fame manner, it may be proved, that the angle CLB is equal to the angle DFE.

The angle MKL, therefore, being equal to the angle DEF, and the angle MLK to the angle DFE, the remaining angle KML will also be equal to the remaining angle EDF; and confequently the triangle KLM is equiangular to the triangle EFD, Q. E. D.

PROP. IV. PROBLEM.

In a given triangle to infcribe a circle,

G

B

Let ABC be the given triangle; it is required to infcribe a circle in it.

Bifect the angles CAB, ABC, with the right lines AD, DB (I. 9.)

Then, fince the angles DAB, DBA are lefs than two right angles (I. 28.), the lines AD, DB, will, if produced, meet each other (I. 25. Cor.)

And, if from the point of intersection D, there be drawn the perpendiculars DF, DG and DE, they will be the radii of the circle required.