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.. Aza is half the cosine-radius.
:: VA, = (half circumradius) + (half cosine-radius)
.. if B,B2, CC, are similarly drawn, then

VA,= V A = VB, = VB = VC,= VC. .. the circle with centre V and radius VA, passes through 4,4,B,B,C,C

Also, since AK, A,A, bisect one another, A_AA_K is a parallelogram.

:. B,C2, C,A2, A,B, are parallels through K to the sides.

This circle is unique, as regards the triangle, in possessing the property of cutting the sides in three concurrent chords parallel to the sides. It is called the Lemoine circle.

203. If, instead of bisecting SK and AK, we divide SK and AK in any the same ratio at two new points V and a, and if the construction of the last article be made, then, by the same reasoning, the six new points A,A,B,B,CC, will lie on a circle.

If, further, K, K, K, are taken on AK, BK, CK at distances from A, B, C equal to twice Aa, BB, Cy respectively, then the figures A,AA,, BBB,K, CCC,K, will be parallelograms. And because Ka, K7, K. divide KA, KB, KC in the same ratio, therefore the lines K,K., K,Ka, KqKy, are parallel respectively to BC, CA, AB.

Hence we have the proposition :

Three parallels to the sides, drawn so as to cut one another in three points Ka, K, K. on the lines KA, KB, KC, will cut the sides in six points lying on a circle.

The circles so described are called Tucker's Circles. They are unique in cutting the sides in six points, which, being joined in pairs, form three chords parallel to the sides.

Observe that the triangles A,B,C, A,B,C, are similar to ABC. For B,A,C2= B,A,C2= A; and so on.

All Tucker's circles have their centres in SK.
In particular;
Taking the centre at K, we have the cosine circle.
Taking the centre at S, we have the circumcircle.

Taking the centre at the middle point of SK, we have Lemoine's circle.

The Brocard Points and the Brocard Angle. 204. PROP. XIV. To draw from the vertices of a triangle three lines intersecting in a point, which shall subtend equal angles at the other vertices.

On AB, BC, CA describe segments of circles containing angles equal to A', B', C" (the supplements of A, B, C) respectively. (Fig. 1.)

These segments will cut in one point, 1 say, because the angles A', B', C' are together equal to four right angles.

Then the exterior angle at 2 of the triangle BAC = B, .. the two interior angles CBA and BCA are together equal to B. .. the angles BCN, ABQ are equal.

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Thus, the three angles ABN, BCN, CAQ are equal.........(1).

Again on CA, AB, BC describe segments of circles containing angles equal to A', B', C" respectively, cutting in N. (Fig. 2.) Then, as before,

the three angles ACO', BAN', CBN are equal.........(2).
If possible, let the angles of (1) and of (2) be unequal.
Suppose those of (1) are the greater.
Then from the triangles ANB, ANC, containing the angle A',

ΑΩ : ΒΩ > ΑΩ' : '.
Similarly ΒΩ : > ΒΩ' : ΑΩ', and CΏ : ΑΩ > ' : ΒΩ'.

But compounding the last two disproportions, we have ΒΩ : ΑΩ > CΏ' : ΑΩ', which contradicts the first disproportion.

.. the six angles of (1) and (2) are equal.

The value, w say, of each of these angles is called the Brocard Angle: and the points 1, 2', the Brocard Points. These points are obviously anticentres.

COR. Turning the above disproportions into proportions, we see that the following pairs of triangles are similar, viz. :

ANB and ANC; BNC and BN'A ; CNA and CA'B. Also since AB : AN=AC : AN', .. the sides about the equal angles ΒΑΩ', CΑΩ of the triangles ΑΩ' Β, ΑΩC are reciprocally proportional : Thus the following pairs of triangles are equal, viz.:

AN'B and ANC; BO'C and BNA; CN'A and CAB.


The Brocard Circle.

t. 205. PROP. XV. The Brocard Points, and three intersectiores of the Brocard Lines lie on the circle whose diameter is the line joining the centres of the Circumcircle and the Cosine Circle.

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Let BO', CO cut in D'; CO2', AN cut in E'; AS2', BN cut in F".

Then ANB, ANC being each the supplement of A, the angle E'N'F" = A and E'QF" is its supplement, .. a circle goes round E'NF'S': similarly a circle goes round D'N'E'N.

.:. the circumcircle of D'E'F' passes through 2, 2.......(1).

Again, the angles BCA, CBN' are equal, .. the perpendicular from D' on BC bisects BC (in D, say).

.. the perpendiculars from D, E, F" on the sides cut in S the circumcentre of ABC.

And since the angle D'SE' (between the perpendiculars on BC, CA)=C, and that D'N'E= the supplement of C, .. S lies on the circle

.(2). Lastly let the diameter through S of this circle cut it in K. Then SDK, SEK, SFK are right-angles.

:. D'K, E'K, FK are parallels to BC, CA, AB.

:. D'D, E' E, F'F are equal to the perpendiculars from K on the sides.

But :: the angles DBD', ECE', FAF are equal, :. the triangles BDD', CEE, AFF are similar; .:. DD' : EE : FF = BD : CE : AF= BC : CA : AB. :: (Prop. VI. Cor.) K is the centre of the cosine circle ... (3).

Cor. I. Since SK is a diameter of the Brocard Circle, the Lemoine and Brocard Circles are concentric.

COR. II. Since KD, KE, KFare parallels to the sides, these meet the sides on the Lemoine Circle.

COR. III. Since KD' is parallel to BC, .. KD'N' = <CBN' = the Brocard angle. Similarly KF'N = the Brocard angle .. KN, KN' subtend equal angles in the Brocard circle, and 20' is bisected at right-angles by SK.

Cor. IV. The triangle D'E' F' is similar to ABC.

General Property of Anticentres. 206. Prop. XVI. If three lines from the vertices of a triangle meet in a point, their antiparallels from the vertices with respect to the vertical angles will meet in a point; and the feet of the perpendiculars from the two points of concurrence on the sides will lie on a circle whose centre is the point midway between the two points of concurrence. J. T.


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