Inversely. The line GH, being perpendicular at the end of the radius CA, it will be a tangent to the circle at A.

Suppose it was not a tangent to the circle, then it would cut the circumference in one more point, as I; join CI, then the triangle CAI, is isosceles by the two radii CA, and CI, and the angles at A, and I, are equal, but the angle at A, is a right angle, therefore, also, the angle at I, is a right angle, and the triangle CAI, has two angles, each equal to one right angle, which is impossible. Therefore, the line GH, has not any other point common with the circumference of the circle, except A, that is, it is a tangent to the circle. As was to be demonstrated.

Corol. At one point in the circumference of a circle, there can be but one straight line tangent to the circle; every other straight line through that point must cut the circle.

Scholium. If a tangent is to be drawn to the circle from a point in the circumference; the radius to that point being drawn, a perpendicular to it at that point, will be the tangent required.

§ 46. PROBLEM. To draw a tangent to a circle, from a given point without it.

APPLIC. The circle EDEG, and the point A, without it, (fig. 50,) being given, it is required to draw from A, a tangent to the circle EDEG.

Solution. Draw AC, joining the centre C, and the given point A, with CA, as radius from the centre C, draw the circumference of the circle, ABFB; at the point D, where the AC, cut the circumference of the given circle, draw EDE, at right angles to AC, it will cut the circumference of the outer circle in B, and B; join the BC, which will cut the circumference of the given circle in E; join AE; and AE, will be the tangent to the given circle from the point A; as was required.

Proof. In the two triangles CAE, and CBD, there is CD=CE, as radii of the given circle, the CB-CA, as radii of the outer circle; the angle at C, is common to both triangles

CBD, and ACE; therefore, these two triangles are equal, and the angles opposite to the equal sides are equal; therefore, the angle AEC, is equal to the angle BDC, being opposite to the equal radii of the outer circle; but BDC, is a right angle by construction, therefore, also, AEC; therefore, the straight line AE, is drawn from the point A, without the circle, perpendicular to the end of the radius CE; therefore this AE, is the tangent required.

Scholium. As every straight line drawn within a circle, cuts the circumference in two points, there will always be two points in the circumference of the outer circle above described, from which radii can be drawn to the common centre of both circles, which will determine two points in the circumference of the given circle; these being joined to the given point, will form tangents to the given circle; and no more are possible, because a straight line can cut the circumference of a circle in no more than two points.

§ 47. THEOREM. In a circle equal chords are equally distant from the centre, smaller chords are at a greater distance, and greater chords at a less distance from the centre; and inversely, chords at equal distances from the centre are equal; chords at a greater distance are smaller; and chords at a less distance are greater; and the diameter is the longest line that can be drawn in the circle.

APPLIC. In the circle, (fig. 51,) let the chords BA, and FE, be equal, their distances CH, and CI, from the centre C, will be equal. The chord AD, less than AB, will have its distance CK, from the centre greater than the distance CH, of the latter. Inversely: the CH, and CI, being equal, the AB, and FE, will be equal; the CK, being greater than CH, the chord DA, will be less than AB. The diameter GL, will be the greatest straight line that can be drawn in the circle.

Demonstr. The perpendiculars CH, CI, and CK, being drawn upon the several chords, join CA, and CE; the perpen

diculars from the centre upon these several chords will bisect them, and the radii of the circle will all be equal; that is, CA=CE=CG=CL; then:

I. The chords AB, and EF, being equal, their distances CH, and CI, are equal; for, the CA, being equal to CE, as radii, if from the square of each of them be taken away the square of the equal half chords AH, and CA there remains in the triangles CHA, and CIE, right angled at H, and I, the square of the third side CH, in the one, equal to the square of the third side CI, in the other, the CH, and CI, are the distances of the chords from the centre, and their squares being equal, the lines are equal; therefore, CH=CI; that is, the equal chords are equally distant from the centre. Inversely: if from the square of the equal radii CA, and CE, there be taken away the square of CH, and CI, respectively, each from each, the differences will be equal; the CH, and CI, being equal by the supposition of equal distances of the chords from the centres; these differences are the squares of the third sides of the right angled triangles CAH, and CEI, the squares being equal, the sides AH, and EI, are themselves equal; therefore, also, their doubles; that is, AB=EF, which are the chords; therefore, at equal distances from the centre, the chords are equal.

II. Let the chord AD, be smaller than AB, then will AD, be farther from the centre C, than AB; that is, the perpendicular CK, from the centre upon AD, will be greater than the perpendicular CH, upon the AB. The radius CA, being common to the two right angled triangles ACH, and ACK, subtracting from the square of AC, as hypathenuse, the square of AH, or of half the chord AB, there remains the square of CH, and subtracting from the same square of CA, the square of AK, there remains the square of CK; but the square of AH, is greater than the square of AK, because the chord AB, is greater than AD, therefore the remaining square CH, is less than the remaining square CK; or CH, is less than CK; that is the distance of the greater chord from the centre is less than

the distance of the smaller chord. Inversely, if from the square of AC, the squares of CH, or of CK, are respectively subtracted, the CK, being larger than the CH, that is, the chord AD, being farther from the centre than the chord AB; thence of the squares remaining by the subtraction in the same right angled triangles as before, the square upon AH, will be greater than the square upon AK, the third sides of these triangles; therefore, also, AH, is greater than AK, therefore also the chord AB, the double of AH, is greater than the chord AD, the double of AK; that is, the chord at the greater distance from the centre is less than the chord nearer to the centre.

III. The diameter is the greatest line that can be drawn in the circle: for AC-EC-GC-LC; therefore, the two parts of the diameter CE, and CL, are each equal to the hypothenuse of any of the right angled triangles used above; their sum LG, is therefore greater than any chord in the circle.

All which was to be demonstrated.

§ 48. THEOREM. In the same or in equal circles, equal angles at the centre subtend equal chords, and equal arcs; and inversely, equal chords, or arcs, subtend equal angles at the centre.

APPLIC. In the same or equal circles EDAGB, and EDF, (fig. 52,) let the angles at the centre C, be ACB= DCE; then will the chords and the arcs AB, and DE, be equal; and inversely, if the chords or arcs AB, and DE, are equal, the angles at the centre C, will be equal.

Demonstr. In the two triangles ABC, and DEC, whether in the same or in equal circles; the sides AC=BC--DC =EC; as radii of the same or equal circles, the angles at C, being equal by the supposition; the third sides AB, and DE, are also equal; that is, the chords are equal. Because, every point in the circumference of the same or equal circles, is equally distant from the centre C, the arc AB, being placed upon the arc DE, will coincide with the same, as will also

the whole sector, having the same angle at the centre, therefore, also, these arcs are equal.

Inversely. If the chords, or the arcs, AB, and DE, are equal, the radii AC, BC, DC, and EC, being all equal, the triangles formed by the equal radii, and the equal chords, or arcs, will also be equal, and coincide with one another in all respects; therefore, also, the angle at the centre C, will be equal; that is, ACB=DCE; as was to be demonstrated.

Corol. The perpendicular from the centre of a circle upon a chord, bisects also the arc subtended by that chord; because it bisects the angle at the centre, and to these equal angles equal arcs will correspond.

§ 49. THEOREM. The angle at the centre of a circle is double the angle at the circumference, that is subtended by the same chord or arc.

APPLIC. In the circles ADBE, (fig. 53 and 54,) let the angles ADB, that has its angular point D, in the circumference, be subtended by the same arc or chord as the angle ACB, at the centre of the circle, the angle ACB, shall be double the angle ADB (This angle is usually said to stand in the segment ADB.)

Demonstr. Join CB, and DC, and protract it to the intersection with the circumference at E, the angle ECB, is outward angle to the triangle BDC; therefore, it is equal to the two inward and opposite angles CBD, and CDB, together; the CD, and CB, being equal, as radii of the same circle, the angles CBD, and CDB, are equal; therefore, the angle ECB, is double the angle EDB; exactly in the same manner is proved, that the angle ACE, is double the angle ADE; therefore, in the two figures 53, the sum of the two angles ACE and BCE, that is, ACB, is double the sum of the two angles ADE, and BDE, that is ADB; and in figure 54, if from the angle EDB, be taken away the angle EDA, and from the ECB, the ECA, the difference between the two last, namely, the angle ACB, will be double the angle ADB;