A treatise of angular sections by John Wallis ...
Wallis, John, 1616-1703., Wallis, John, 1616-1703. Treatise of algebra.
Page  50

### CHAP. VI. Of the Proportion of the Base to the Legs of a Triangle, according as is the Angle at the Top of it.

I. THE noted Proposition of Pythagoras, (which is in Euclid, 47 è 1.) concerning the Square of the Base equal to the Squares of the two Legs containing a Right-angle: And two more in Euclid (Pr. 12, 13, è 2.) concerning the Excess, (in case the Angle at the Top be Obtuse;) or the Defect, (in case it be Acute;) of the Square of the Base, compared with the Squares of the two Legs: And some other Propositions in the foregoing Chapters, shewing what Proportion that Excess or Defect bears to a Rectangle of the Legs, in divers cases: Gave me occasion to pursue that Spe∣culation a little further; according to the following Propositions.

[illustration]

II. If by the Legs of a Triangle C, D, the Angle at the Top contained A, be a Right-angle, (or of 90 De∣grees;) and from thence a Perpendicular G, on the Base, cut this into two Segments χ, δ: The two Triangles hence arising, χ G C, G δ D, are like to the whole CDB; (because of one Angle common, the other a Right∣angle, and therefore the third equal to the third.) And therefore (the Triangles being here designed by their sides.)

〈 math 〉 and therefore 〈 math 〉 and therefore, Cq+Dq=(Bχ+Bδ=B: into χ+δ=B:=) Bq. That is,

The Square of the Base is equal to the two Squares of the Legs containing a Right-angle.

[illustration]

III. If the Angle A be 120 Deg. or 4/3 of a Right∣angle; and thence be drawn the Base two streight Lines G, Γ; making, with it, Angles equal to that at the Top: The Two Triangles χ GC, ΓδD, are like to the whole CDB, (because of one Angle common, another equal to A, and therefore the third equal to the third;) with a Triangle between GΓμ Equi∣later (because each of the Angles at the Base, and therefore that at the Top, are of 60 Degrees:) And the Base B=χ+δ+μ. And therefore, 〈 math 〉 and 〈 math 〉 Therefore, Cq+Dq+CD= (Bχ+Bδ+Bμ=B: into χ+δ+μ=) B:=Bq. That is,

The Square of the Base (of an Angle of 120 Degrees) is equal to the Squares of the Legs and a Rectangle of them.

Page  51IV. If the Angle A be 60 Degrees, or ⅔ of a Right∣angle; and from thence, G, Γ, making Angles with the Base equal to that of A: Then are, (for the same causes as before) χGC, ΓδD, like Triangles to CDB, (as before;) and GΓμ an Equilater Triangle com∣municating with them: And the Base, B=χ+δ−μ. (or χ−μ+δ.) And therefore,

[illustration]
[illustration]
〈 math 〉 and 〈 math 〉 And therefore, Cq+Dq−CD=(Bχ+Bδ−Bμ=B: into χ+δ−μ=B=) Bq. That is,

The Square of the Base (of an Angle of 60 Degrees) is equal to the Squares of the Legs, wanting the Rectangle of them.

(Note here, that, by χ, I understand the Base to the Legs CG; by δ, that of the Legs DΓ; by μ, that of GΓ; and, by B, that of CD, which is ever equal to χ+δ±μ, however these parts be intermingled. Which where it is +μ, is commonly more obvious to the Eye; but where it is −μ, is more perplex, and will need more consideration to discern; but it is equally true in both cases.)

V. If A be 135 Degrees (or 3/2 of a Quadrant,) and GΓ drawn (as before) to make the like Angle with the Base: The Triangles χGC, Γ δ D, will be like to CDB; and GΓμ will be Equi∣crural,

[illustration]
making the Angles at the Base, of 45 De∣grees, (so much as A wants of two Right-angles▪) and therefore, the Angle at the Vertex (which I shall call V,) of 90 Degrees. And therefore, (by § 2.) μq=2Gq, and 〈 math 〉 And the Base B=χ+δ+μ. And therefore, 〈 math 〉 and 〈 math 〉. Therefore, 〈 math 〉 into χ+δ+μ=B:=) Bq. That is,

The Square of the Base (of an Angle of 135 Degrees) is equal to the Squares of the Legs, with a Rectangle of them Multiplied into〈 math 〉

[illustration]

VI. If A be 45 Degrees: It will in like manner be shewed, that (because of B=χ+δ−μ.) 〈 math 〉 into χ+δ−μ=B:=) Bq. That is,

The Square of the Base (of an Angle of 45 Degrees) is equal to the Squares of the Legs, wanting a Rectangle of them Multiplied into〈 math 〉

Page  52VII. And, universally, what ever be the Angle A; it will (by like process) be shewed: That,

〈 math 〉. That is, The Square of the Base (whatever be the Angle at the Vertex) is equal to the Squares of the Legs, together with (if it be greater than a Right-angle) or wanting (if less than such) a Plain, which shall be, to the Rect-angle of the Legs, as a Portion in the Base-line, intercepted between two Lines from the Vertex, making at the Base a like Angle with that of the Vertex, to one of those two Lines so drawn.

VIII. Of this we are to give great variety of Examples in the following Chap∣ter, where this General Theorem is applied to particular cases: And which is further improved by these two ensuing Propositions.

[illustration]

IX. The Radius of a Circle, with the subtenses of two Arches, being given; the Subtense of their Aggregate is also given. For, supposing the subtenses of the given Arches to be A, E: The subtenses of their Remainders to a Semicircle, are also had: Suppose 〈 math 〉 And 〈 math 〉. And therefore, inscribing a Quadrilater whose opposite sides are A, ε; and E, α; one of the Diagonals is the Dia∣meter=2R; the other the Subtense of the Sum or Ag∣gregate of those Arches, suppose 〈 math 〉.

[illustration]

X. The same being given; the Subtense of the Difference of those Arches is also given. For, having (as before) A, α; E, ε; 2R: we have (by a Quadrilater duly inscribed) the Subtense of the Difference, 〈 math 〉.

XI. It is manifest also, (from what is before delivered,) that the same Tri∣angle GΓμ, doth indifferently serve for the Angle of 120 Degrees and of 60 Degrees: And, in like manner, for 135, and 45: And so, for any two Arches whereof one doth as much exceed as the other wants of a Quadrant. For, the Angle V is in both the same; and the Angles at the Base differ only in this: That, in one, the External Angle; in the other, the Internal, (which is the others Complement to two Right-angles;) is equal to the Angle of CD at the Vertex.

XII. Hence it follows: That, Of two Angles, where the Legs of the one are respectively equal to those of the other; the one as much exceeding a Right-angle, as the other wants of it: The Square of the Base in the one, doth as much exceed the two Squares of the Legs; as, in the other, it wants thereof.

[illustration]

XIII. And consequently, In any Right-lined Triangle, (however inclined,) the Squares of the Axis or Diameter, and of the half Bases twice taken; are equal to the Squares of the Legs. For, supposing C, C, the two halfs of the Base; and B, the Diameter or Axis of the Triangle, (meaning thereby a streight Line from the Vertex to the middle of the Base;) and B, β, the two Legs: It is manifest, that, of the two Angles at the Base (which are each others Complement to two Right-angles;) the one doth as much exceed, as the other wants of, a Right-angle: And there∣fore the Square of one of the Legs, as Bq, doth as much exceed; as the other, βq, doth come short of; Dq+Cq. And therefore, both together, Bq+βq=2Dq+2Cq.

Page  53XIV. And therefore, The Base, and Axis (or Diameter) of a Triangle remain∣ing the same; (however differently inclined:) the Aggregate of the Squares of the two Legs, remains the same.

XV. And the same is to be understood of the Squares of Tangents, of a Parabola, Hyperbola, Elipsis, (or other Curve Line having Diameter and Ordinates,) from the two ends of an Inscribed Ordinate, to the Point of the Diameter (produced if need be) wherein those Tangents meet.

XVI. The same may be likewise accommodated, to the Segments (of such Legs, or Tangents,) Cut off by Lines Parallel to the Base. Namely, The Squares of such Segments (intercepted by those Parallels) together taken, (the Axe of such Tra∣pezium remaining the same,) are the same: Whether such Trapezium be Erect, or however inclined. For such Segments, are still Proportional to their Wholes.