Cor. 1.-Hence, if from any point without a circle, two right lines be drawn cutting it, the rectangles under these right lines and their external segments are equal, being each equal to the square of a tangent, drawn from the same point to the circle.

Cor. 2.-Two tangents drawn to a circle, from any point without it, are equal.

For their squares are equal, being each equal to the same rectangle.

Cor. 3. From this, and the preceding proposition, it is manifest, that, if a right line, passing through any point, either within or without a circle, cut it in two points or touch it; the square of the segment of the tangent, or rectangle under the segments of the secant, between that point, and the point or points, in which it touches or cuts the circle, is equal to the difference of the squares of the radius, and the distance of the same point from the centre of the circle.

Cor. 4.-Hence also if two right lines, meeting each other, both touch, or both cut, or one of them touch, and the other cut a circle; the squares of the segments of the tangents, or rectangles under the segments of the secants or cutting lines, between their concourse, and the points in which they meet the circle, are equal.

PROP. XXXVII. THEOR.

If from any point (E) without a circle, two right lines (DE, AE) be drawn, one of them (DE) cutting the circle, and the other (AE) meeting it, and the rectangle (DEC) under the cutting line, and its external segment, be equal to the square of the line which meets it; the right line (AE) which meets the circle, touches it.

From E, draw EB touching the circle (17. 3), find the centre F (1. 3), and join FA, FE, FB.

Because EB touches the circle, and DE cuts it, the square of EB is equal to the rectangle DEC (36. 3); but the square of AE is equal to the rectangle DEC (Hyp.), therefore the squares of AE and EB are equal (Ax. 1. 1), and, of course, the right lines AE, ÉB them- D selves (Cor. 1. 46. 1); therefore, in the triangles FAE, FBE, the sides FA, AE

A

F

E

are severally equal to FB, BE, and FE is common to both triangles, therefore the angles FAE, FBE are equal (8. 1); but FBE is a right angle (18. 3), therefore FAE is also a right angle, and, of course AE touches the circle (Cor. 16. 3).

BOOK IV.

DEFINITIONS.

1. A RECTILINEAL figure, is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure, are in the perimeter of the other.

2. A rectilineal figure, is said to be circumscribed about another rectilineal figure, when the perimeter of the former touches all the angles of the other. 3. A rectilineal figure, is said to be inscribed in a circle, when all its angles, are in the circumference of the circle.

4. A circle, is said to be circumscribed ǎbout a rectilineal figure, when every angle of the rectilineal figure, is in the circumference of the circle.

5. A rectilineal figure, is said to be circumscribed about a circle, when every side of it, touches the circle.

6. A circle, is said to be inscribed in a rectilineal figure, when every side of the rec tilineal figure, touches the circle.

7. A regular figure, is that, which is equilateral and equiangular.

PROPOSITION I. PROBLEM.

In a given circle (CAB), from a given point (C) in its circumference, to inscribe a right line, equal to a given right line (D), not greater than the diameter of the circle.

Draw the diameter of the circle CB, and, if this be equal to D, what was required is done.

If not, take from CB, a part CE equal to D (3. 1), and from the centre C, at the distance CE, describe the cir

cle AEF, and to either of its DL

intersections with the given circle, as A, draw CA, this is equal to CE (Def. 10. 1), and therefore to the given right line D (Constr. and Ax. 1. 1).

Cor.-Hence it appears, how, in a given circle, from a given point in its circumference, an arch may be taken, equal to a given arch, of an equal circle; namely, by drawing the chord of the given arch, and inscribing in the given circle, from the given point, a right line equal to that chord (by this prop.), which right line cuts off an arch equal to the given one (by 28. 3).

PROP. II. PROB.

In a given circle (BAC), to inscribe a triangle, equiangular to a given triangle (EDF).

Draw the right line GH, touching the circle in any point A (17. 3), and at the point A, with the right line AH, make the angle HAC equal to the angle E (23. 1); and at the same point, with the right line AG, make the angle GAB equal to angle F, and join BC.

AD

E

T

The angle E is equal to the angle HAC (Constr.), or, (32. 3), to the angle B in the alternate segment; for à like reason, the angles F and C are equal; therefore the remaining angle D is equal to the remaining BAC (32. 1); therefore the triangle BAC, which is inscribed in the given circle (Def. 3. 4) is equiangular to the given triangle EDF.

PROP. III. PROB.

About a given circle (ABC), to circumscribe a triangle, equiangular to a given triangle (EDF),

Prod e any side EF, of the given triangle, both ways, as to G and H; find the centre K of the given circle (1. 3), from which draw any radius KA, with which, at K, make the angle BKA equal to DEG (23. 1); and, with BK at K, the angle BKC equal to DFH; and draw ML, MN and LN, touching the circle in the points A, B and C (17. 3).

Because the four angles, of the quadrangle MAKB, are equal to four right angles (Cor. 1. 32. 1), and the angles KAM, KBM are right angles (18. 3), the remaining angles AKB, AMB are equal to two right angles, and therefore to DEF, DEG, which together are also equal to two right angles [13. 1]; taking away the equal angles AKB, DEG, the remaining angles AMB, DEF are equal: in like manner, LNM and DFE may be proved equal; therefore the remaining angle L is equal to the remaining D [32. 1], and, of course, the triangle LMN, which is circumscribed about the given circle [Def. 5. 4], is equiangular to the given triangle DEF.

PROP. IV. PROB.

In a given triangle (ABC), to inscribe a circle.

D

Bisect any two angles ABC, BCA of the given triangle by the right lines BD, CD (9. 1), which, because the angles DBC, BCD, are together less than ABC, BCA together [Ax. 9], and therefore than two right angles [17. 1], may be so produced, as to meet (Theor. at 29. 1); let them be so produced, and meet, as in D, from which, let fall the perpendicular DE on AB [12. 1], from the centre D, at the distance DE, describe a circle (Post. 3), which is inscribed in the given triangle.

B

For, the perpendiculars DF and DG being let fall on BC and CA; because, in the triangles DEB, DFB, the angles DEB, DBE are severally equal to DFB, DBF [Constr.], and the side DB common, DE and DF are equal [26. 1]; in like manner, DF, DG may be proved equal; therefore the three right lines DE, DF, DG are equal [Ax. 1. 1], and the circle, described from the centre D, at the distance DE, passes through F and G; and since the angles formed by AB, BC, CA with the radiuses, at the points E, F, G are right angles, these right lines touch the circle (Cor. 16. 3), which is therefore inscribed in the triangle ABC [Def. 6. 4), as was required,