FOURTH BOOK. THE PROPORTIONS OF LINES AND THE AREAS OF FIGURES IN CONNECTION WITH THE CIRCLE. DEFINITIONS. I. Similar arcs, Sectors, or Segments, are those which, in different circles, correspond to equal angles at the centre. II. When two similar sectors are superposed so that their equal angles coincide, their difference is called a circular trapezoid. Thus the space BB'C'C is a circular trapezoid. III. The space included between two concentric circumferences is called a circular ring. IV. The arc of a circle is said to be rectified, when it is developed; that is, unfolded or drawn out into a straight line. It is only in this rectified form that we can conceive of its length, since it could not otherwise be compared with the linear unit. PROPORTIONAL LINES CONNECTED WITH THE CIRCLE. THEOREM I. When two chords intersect each other within a circle, the segments will be reciprocally or inversely proportional. A B E Let the two chords AA', BB' intersect at P. Drawing the auxiliary chords AB', A'B, we thus obtain two triangles PAB', PA'B, which are similar, since the angles at P are equal (B. I., T. I.); the angles at A and B are equal, so also are the angles at A' and B' (B. II., T. X.), hence these triangles are equiangular, and consequently similar (B. III., T. VI.), and their homologous sides give this proportion, PA: PB:: PB': PA'. Scholium I. When one of the chords is a diameter, and the other is perpendic ular to it, we have PA: PB:: PB: PA'. The line PB, drawn perpendicular to a diameter, is called an ordinate to this di ameter. Hence, we have this condition: B In a circle any ordinate to a diameter is a mean proportional between the two segments of this diameter. We can show that this property is one of the consequences of the properties of right-angled triangles; because if we join A, A' with the point B, we shall form a right-angled triangle (B. II., T. X., C. I.), which gives (B. III., T. XIII.) this proportion, PA: PB:: PB: PA'. Scholium II. This same right-angled triangle ABA' gives the proportion, That is, AA': AB::AB: AP. Any chord which is drawn through the extremity of a diameter is a mean proportional between its projection (B. III., T. XIV., S. II.) on the diameter and the diameter itself. THEOREM II. When two secants intersect each other, without a circle, they will be reciprocally proportional to their external segments. Let the two secants intersect each other at P. Draw, as in the preceding Theorem, the chords AB', A'B, and the two triangles AB'P, BA'P, will be similar, since the angle at P is common, and A=B, each being measured by half of the arc A'B' (B. II., T. X.), and the homologous sides give the proportion P A' B' E B PA: PB:: PB': PA'. THEOREM III. When a secant and tangent are drawn from the same point, the tangent is a mean proportional between the secant and its external segment. This Theorem is in reality only a particular case of the preceding, when the points B, B' of the secant PB, are united in one point. But we will give a direct demonstration, by drawing the chords AB, A'B. We have, in effect, two triangles, PAB, PA'B, similar, since the angle at P is common and the two angles at A and B are equal (B. II., T. X., C. II.). Hence, we have this proportion, A' P E B PA: PB:: PB: PA'. When a straight line, as PA, is thus divided at a point A', so that the greater segment AA', is a mean proportional between the whole line and the other segment, it is said to be divided into mean and extreme ratio. or, The above proportion will give AA': PA-AA':: PA': AA'- PA', AA': PA':: PA': AA'- PA'. Now if we take PA" equal to PA', and observe that AA'= PB, and consequently that AA'-PA'= PB-PA"= A′′B, we shall have PB: PA":: PA": A"B. From this we see that the straight line PB is also divided at the point A" into mean and extreme ratio. Scholium II. The three preceding Theorems give immediately these two equations, PA× PA'=PB × PB', PA× PA'=PB2, which may be included in one single proposition, as follows: The product of the distances from the same point, either within or without a circle, to two points of its circumference, taken in the same straight line, is always the same. In the case of the tangent we must regard the two points of the circumference as united. THEOREM IV. In a quadrilateral inscribed in a circle, the rectangle of the diagonals is equal to the sum of the rectangles of the opposite Let ACBD be the inscribed quadrilateral, and we shall have ABX CDAC x BD+AD × BC. For, drawing CE, making the angle ACE equal to the angle BCD, and consequently the angle ECB equal to the angle ACD. B E C' The two triangles CEA, CBD are similar, since the angle ACE equals the angle DCB, by construction, and CAE = CDB, since each is in the same segment (B. II., T. X.). Their homologous sides give this proportion, consequently, AC: CD:: AE: BD; CD × AE = AC × BD. The comparison of the homologous sides of the two triangles BEC, DAC, which are similar, gives and AD: BE:: CD: BC, CD × BE AD × BC. = Adding the corresponding members of these equations we obtain or CD × (AE+BE) = AC × BD+AD × BC, Scholium. This Theorem has many important applications. FIRST. To find the chord AB of the sum of two arcs AC, CB, when their chords are known. For abridgment, let a, b, c denote the three sides BC, AC, AB of the triangle ABC, and R the radius of its circumscribing circle. Draw the diameter COC'= 2R, also the supplementary chords AC', BC'. Now the inscribed quadrilateral ACBC' will give AB × CC'=AC × BC'+ AC' × CB. But the triangles CAC', CBC' being right-angled, since they are each in a semicircle, give AC′= √CC” — AC2, BC= √CC'2 — BC2; and since CC'=2R, AC=6, BC= a, the foregoing expressions will become AC=√4R2 — b2, BC'=√4R2 — a2, |