A treatise of angular sections by John Wallis ...
Wallis, John, 1616-1703., Wallis, John, 1616-1703. Treatise of algebra.

### CHAP. IX. Of Angles compared with they Arches on which the stand.

I. THAT The Angle of a Sector, is proportional to the Arch on which it doth insist; whether such Angle be at the Center, or at the Circumference: And, that such Angle at the Center, is double to that at the Circumfe∣rence: Is shewed by Euclide long since; and is generally known. But not so, in case such Angle be any where else, whether within or without the Circle. It will therefore be not amiss, to pursue that Notion a little far∣ther; as here followeth.

[illustration]

II. If on the Diameter of a Circle D F, be for∣med, at the Center C, an Angle B A D; the inter∣cepted Arch B D, is Proportional to the Angle. (by 26, 27. El. 3. and 33, El. 6, of Euclid.) That is, the Arch intercepted B D, is such a Part of the whole Circumference; as is the Angle A, of Four Right-angles. And accordingly the Angle A, is said to be, of so many Degrees, as is the Arch B D.

III. If B A, D A, the Legs containing such Angle, after a Decussation at C, forming the Vertical Angle E, be continued, on the other side, to the Circum∣ference; the intercepted Arch G F, will, (by § 2.) be equal to B D, (because of the Vertical Angle E, equal to A:) And, consequently, the Aggregate of both B D + G F, is double to B D. That is, B D + G F = 2 B D.

IV. If at F the end of the Diameter, be formed a like Angle A; the inter∣cepted Arch H D, is likewise double to B D. That is, H D = B D + G F = 2 B D. Because (by 20. El. 3.) the Angle at the Center, is double to that at the Circumference, on the same Arch.

Page  67Or thus, because (by Construction) F H, G C B, are Parallels: (as making equal Angles with F D:) Therefore, H B = F G = B D. And, consequently, H D = H B + B D = F G + B D = 2 B D. (By § 3.)

V. If at K, (any other Point of the Diameter within the Circle,) such Angle A be made, (with its Vertical E:) the Aggregate of the two Arches intercep∣ted, F L + M D, is Double to B D. That is, F L + M D = F G + B D = 2 B D. For, (drawing the streight Line A M,) the two Internal Angles F M L + M F D, are equal to the External M K D = A = H F D (which is an Angle at the Circumference.) And therefore, the Arches opposite to those F L + M D, equal to the Arch opposite to this H D = 2 B D. (By § 4.)

Or thus, because F H, L M, be Parallels (as making like Angles with F D,) therefore, F L = H M, and F L + M D = H M + M D = H D = 2 B D. (By § 4.)

VI. If at N, a Point of the Diameter produced, without the Circle; be formed a like Angle A; the Difference of the two intercepted Arches, Q D − P F, is equal to the same H D, or the Double of B D. For, P Q, F H, being Parallels, (as making like Angles with D F produced,) and therefore, P F = Q H: Therefore, Q D − P F = Q D − Q H = H D = 2 B D. (By § 4.)

VII. The same will hold, though neither of the Legs containing the Angle do pass through the Center, (and therefore lie not upon a Diameter;) as I shall now shew in the several Cases.

VIII. If M L, μ λ, make M K μ an Angle at K any where, within the Circle, Let B c β be a like Angle at the Center, and the Legs of this Parallel to those of that. And, by the Angular Point K, draw the Diameter F K D. Then is (by § 5.) M D + L F = 2 B D: And μ D + λ F = 2 β D. Therefore, (the Sum or Difference) M μ (= M D ± μ D) + L λ (= L F ± λ F) = (2 B D ± 2 β D =) 2 B β.

[illustration]

IX. In like manner: If H F η be an Angle at the Circumference: And, at the Center, B C β like to it, and with Legs Parallel to those: And F D a Di∣ameter. Then (by § 4.) H D = 2 B D, and η D = 2 β D. Therefore, H η (= H D ± η D) = (2 B D ± 2 β D =) 2 B β.

[illustration]

Page  68X. The same is to be understood, in case one of the Legs touch the Circum∣ference at F, (the Points F, H, being in this case Co-incident; the Arch F H vanishing to nothing, and the Arch intercepted H η, the same with F η.) For here also, H F D = 2 B D, and η F D = 2 β D; and therefore, H F η (= H F D ± η F D) = (2 B D ± 2 β D =) 2 B β.

[illustration]

XI. In like manner: If Q N χ be an Angle without the Circle, whose Legs cut it in P, π: And, at the Center, a like Angle and like sited B C: And N F D a Diameter produced. Then (by § 6.) Q D − P F = 2 B D, and χ D − π F = 2 β D. And therefore Q D ± χ D wanting P F ± π F, that is Q χ − P π, is equal to 2 B D ± 2 β D = 2 B β.

[illustration]

XII. The same is to be understood, in case one or both of the Legs do not cut, but only touch the Circle. For then the Points P, Q; or π, χ; (or both,) being Co-incident; the rest proceeds as before. For still Q D − P F = 2 B D, and χ D − π F = 2 β D; and therefore, (the Sum or Difference) Q χ − P π = 2 B β.

[illustration]

XIII. But in case one or both of the Legs pass by the Circle, and neither cut, nor so much as touch it: It doth not concern the present business; for such Angle doth not insist on a Circular Arch. The whole therefore (thus demonstrated) amounts to this General.

XIV. If a Circle be cut (or at least touched) by two streight Lines, making an Angle: (and so, when continued, intersecting each other:) The Sum (if their intersection be within the Circle,) or Difference (if without,) of the two Arches intercepted by them (produced, if need be,) or (if their intersection be at the Cir∣cumference) the single Arch by them intercepted; is Double to the Arch of a like Angle at the Center.