Commentary on Proposition VI.3

Back to Proposition VI.3.

The demonstration assumes that CE will meet BA produced in some point E. This is proved in the same way as it is proved in VI. 4 that BA, ED will meet if produced. The angles ABD, BDA in the figure of VI. 3 are together less than two right angles, and the angle BDA is equal to the angle BCE, since DA, CE are parallel. Therefore the angles ABC, BCE are together less than two right angles; and BA, CE must meet, by I. Post. 5.

The corresponding proposition about the segments into which BC is divided externally by the bisector of the external angle at A when that bisector meets BC produced (i.e. when the sides AB, AC are not equal) is important. Simson gives it as a separate proposition, A, noting the fact that Pappus assumes the result without proof (Pappus, VII. p. 730, 24).

The best plan is however, as De Morgan says, to combine Props. 3 and A in one proposition, which may be enunciated thus: If an angle of a triangle be bisected internally or externally by a straight line which cuts the opposite side or the opposite side produced, the segments of that side will have the same ratio as the other sides of the triangle; and, if a side of a triangle be divided internally or externally so that its segments have the same ratio as the other sides of the triangle, the straight line drawn from the point of section to the angular point which is opposite to the first mentioned side will bisect the interior or exterior angle at that angular point.

B D C F A E B C D E A F

Let AC be the smaller of the two sides AB, AC, so that the bisector AD of the exterior angle at A may meet BC produced beyond C. Draw CE through C parallel to DA, meeting BA in E.

Then, if FAC is the exterior angle bisected by AD in the case of external bisection, and if a point F is taken on AB in the figure of VI. 3, the proof of VI. 3 can be used almost word for word for the other case. We have only to speak of the angle “FAC” for the angle “BAC,” and of the angle “FAD” for the angle “BAD” wherever they occur, to say “let BA, or BA produced, meet CE in E,” and to substitute “BA or BA produced” for “BAE” lower down.

B D C E A F

If AD, AE be the internal and external bisectors of the angle A in a triangle of which the sides AB, AC are unequal, AC being the smaller, and if AD, AE meet BC and BC produced in D, E respectively, the ratios of BD to DC and of BE to EC are alike equal to the ratio of BA to AC.

Therefore BE is to EC as BD to DC,

that is, BE is to EC as the difference between BE and ED is to the difference between ED and EC, whence BE, ED, EC are in harmonic progression, or DE is a harmonic mean between BE and EC, or again B, D, C, E is a harmonic range.

Since the angle DAC is half of the angle BAC, and the angle CAE half of the angle CAF, while the angles BAC, CAF are equal to two right angles,

the angle DAE is a right angle.

Hence the circle described on DE as diameter passes through A.

Now, if the ratio of BA to AC is given, and if BC is given, the points D, E on BC and BC produced are given, and therefore so is the circle on D, E as diameter. Hence the locus of a point such that its distances from two given points are in a given ratio (not being a ratio of equality) is a circle.

This locus was discussed by Apollonius in his Plane Loci, Book II., as we know from Pappus (VII. p. 666), who says that the book contained the theorem that, if from two given points straight lines inflected to another point are in a given ratio, the point in which they meet will lie on either a straight line or a circumference of a circle. The straight line is of course the locus when the ratio is one of equality. The other case is quoted in the following form by Eutocius (Apollonius, ed. Heiberg, II. pp. 180—4).

Given two points in a plane and a proportion between unequal straight lines, it is possible to describe a circle in the plane so that the straight lines inflected from the given points to the circumference of the circle shall have a ratio the same as the given one.

Apollonius’ construction, as given by Eutocius, is remarkable because he makes no use of either of the points D, E. He finds O, the centre of the required circle, and the length of its radius directly from the data BC and the given ratio which we will call $h : k$. But the construction was not discovered by Apollonius; it belongs to a much earlier date, since it appears in exactly the same form in Aristotle, Meteorologica III. 5, 376 a 3 sqq. The analysis leading up to the construction is, as usual, not given either by Aristotle or Eutocius. We are told to take three straight lines $x$, CO (a length measured along BC produced beyond C, where B, C are the points at which the greater and smaller of the inflected lines respectively terminate), and $r$, such that, if $h : k$ be the given ratio and $h > k$,

$k : h = h : k + x$, $(\alpha)$ $x :$ BC $= k :$ CO $= h : r$. $(\beta)$

B D C O (E) P L

This determines the position of O, and the length of $r$, the radius of the required circle. The circle is then drawn, any point P is taken on it and joined to B, C respectively, and it is proved that

PB $:$ PC $= h : k$.

We may conjecture that the analysis proceeded somewhat as follows.

It would be seen that B, C are “conjugate points” with reference to the circle on DE as diameter. (Cf. Apollonius, Conics, I. 36, where it is proved, in terms, for a circle as well as for an ellipse and a hyperbola, that, if the polar of B meets the diameter DE in C, then EC $:$ CD $=$ EB $:$ BD.)

If O be the middle point of DE, and therefore the centre of the circle, D, E may be eliminated, as in the Conics, I. 37, thus.

Since

EC $:$ CD $=$ EB $:$ BD,

it follows that EC $+$ CD $:$ EC $\sim$ CD $=$ EB $+$ BD $:$ EB $\sim$ BD, or

$2$ OD $: 2$ OC $= 2$ OB $: 2$ OD,

that is,

BO $\cdot$ OC $=$ OD2 $= r^2$, say.

If therefore P be any point on the circle with centre O and radius $r$,

BO $:$ OP $=$ OP $:$ OC,

so that BOP, POC are similar triangles.

In addition, $h : k =$ BD $:$ DC $=$ BE $:$ EC

$=$ BD $+$ BE $:$ DE $=$ BO $: r$.

Hence we require that

BO $: r = r :$ OC $=$ BP $:$ PC $= h : k$. $(\delta)$

Therefore, alternately,

$k :$ CO $= h : r$,

which is the second relation in $(\beta)$ above.

Now assume a length $x$ such that each of the last ratios is equal to $x :$ BC, as in $(\beta)$.

Then

x : BC = k : CO = h : r.

Therefore

x + k : BO = h : r,

and, alternately,

x + k : h = BO : r = h : k, from (delta) above;

and this is the relation $(\alpha)$ which remained to be found.

Apollonius’ proof of the construction is given by Eutocius, who begins by saying that it is manifest that $r$ is a mean proportional between BO and OC. This is seen as follows.

From $(\beta)$ we derive

x : BC = k : CO = h : r = (k + x) : BO,

whence

BO : r = (k + x) : h = h : k, by (alpha), = r : CO, by (beta),

and therefore

r^2 = BO ,., CO.

But the triangles BOP, POC have the angle at O common, and, since BO : OP = OP : OC, the triangles are similar and the angles OPC, OBP are equal.

[Up to this point Aristotle’s proof is exactly the same; from this point it diverges slightly.]

If now CL be drawn parallel to BP meeting OP in L, the angles BPC, LCP are equal also.

Therefore the triangles BPC, PCL are similar, and

BP : PC = PC : CL,

whence

BP2 : PC2 = BP : CL = BO : OC, by parallels, = BO2 : OP2 (since BO : OP = OP : OC).

Therefore

BP : PC = BO : OP = h : k (for OP = r).

[Aristotle infers this more directly from the similar triangles POB, COP. Since these triangles are similar,

OP : CP = OB : BP,

whence

BP : PC = BO : OP = h : k.]

Apollonius proves lastly, by reductio ad absurdum, that the last equation cannot be true with reference to any point P which is not on the circle so described.