33. Theorems I. n is the common positive Brocard point of ABC and A'B'C'.

Since ACA' is a cyclic quadrilateral NAB = NC'A' = w (Euc. III. 21); similarly QB'C' and QA'B' are each equal to w.

It follows also that ' is the common negative Brocard point of ABC and A"B"C".

II. The sides of A'B'C' and A"B"C" are equally inclined to the corresponding sides of ABC.

For by (1) CB'C'′ = AC'A' = BA'B' = 90 — w,

and

BC"B" AB"A" CA"C" = 90-w.

=

=

III. The six points A', B, C, A", B", C" are concyclic. For the angles AC′A′ = AB′′A′′, therefore A'A"B"C′ is cyclic (Euc. III. 22).

But

Similarly B'B"C"A' and C'C"A"B' are cyclic. generally if three pairs of points on the sides of a triangle are such that every two pairs are cyclic, the six points lie on a circle.* For if they do not the tangents to the three circles from A, B and C are easily seen to be equal, which is impossible.

IV. The lines B"C', C"A', A"B' are parallel to the sides a, b, c respectively.

We know that each pair of sides of ABC with N and

form similar triangles, i.e., BNC and AN'C, CNA and ΒΩ'Α, ΑΩΒ and CSB are similar; hence the perpendiculars (or other corresponding lines) through N and ' divide the opposite sides similarly. In the triangles CNA

* For example, if A'B'C' be the middle points of the sides and A"B"C" the feet of the perpendiculars, it follows immediately that A'B'C'A"B"C" is a cyclic hexagon since each pair of points AA' and BB′ form a cyclic quadrilateral. ("Nine Points" Circle.)

and BOA we have therefore AC AC-AB" AB, or B"C'

is parallel to a.

V. Hence also A'A", B′B", C'C" are antiparallels to the sides a, b, c. (Euc. III. 22.)

SECTION III.

TUCKER'S CIRCLES.

34. By Art. 24 if the inscribed triangle A'B'C' is given in species only it may be conceived to vary its position by rotating around the point which is fixed. Let it revolve in a positive direction through any angle and also let A"B"C" revolve in the opposite direction through an equal angle.

Then each of the equal angles of inclination of the sides of A'B'C' and A"B"C" are diminished by e, therefore for all values of the sides are equally inclined and the vertices of the two triangles are always concyclic.

The circles thus described are called the Tucker Circles of the triangle.

Thus the lines B'C' and A'A", etc., are always parallel and antiparallel respectively to the opposite side a, and therefore remain constant in direction.

Now since the point N is fixed and the triangle A'B'C' of constant species; since the vertices move on given lines all points fixed relatively to the figure describe lines. The locus of the centre of the system of Tucker's circles is therefore a line. (Art. 20.)

By taking particular positions of the triangle we find points on the line of centres. In the case where 0=0 the

vertices of ABC and A'B'C' coincide, and the circumcircle is thus seen to be one of Tucker's circles. The line of centres thus passes through the circum-centre of ABC. Similarly the loci of the other Brocard points of the triangle A'B'C' and A"B"C" are lines.

35. Let the vertices of the triangle formed by the parallels B'C', C"A', A"B' to the sides of ABC be denoted by X, Y, Z.

Then AA'A"X is a parallelogram, as are also BB'B"Y, CC'C"Z; and since the diagonals bisect each other AX bisects the antiparallel A'A". AX, BY, CZ are the symmedians of ABC.

Hence the following construction for Tucker's circles:

Let K be the symmedian point of ABC. Join AK, BK, CK. Take any point X on AK and draw parallels through it to the sides b and c. Let them meet BK and CK in Y and Z respectively. YZ is parallel to a, and the hexad of points in which the sides of ABC are cut by these parallels lie on one of the required circles.

36. The antiparallels A'A", B'B", C'C" are equal. For since A′′B' is parallel to c, and A'A" and B'B" are equally inclined to c (at an angle C), A'A"=B′B" ; therefore, etc.; or they are the chords of a Tucker circle intercepted by parallel lines.

37. Theorem. The line OK is the locus of the centre of Tucker's system of circles.

For let L be the middle point of the chord A'A" of one of the system. Draw LO1 at right angles to it meeting OK in 01. Join AO.

Since the tangent at A to the circum-circle is antiparallel to a, AO and LO1 are parallel lines.

1

But AK|AX=BK|BY=CK/CZ (Euc. VI. 2); therefore AK|AL=BK|BM = CK/CN= OK/001, or O̟, is the centre of the Tucker circle.

38. Since is the positive Brocard point of the triangles ABC and A'B'C', and NAB and NA'B' a pair of similar triangles; if be the inclination of the sides of A'B'C' to those of ABC, we have

This ratio is the Ratio of Similitude of the triangles, and is the constant relation between all corresponding lines of A'B'C' and ABC.

be the radius of Tucker's circle for

For example, if

In (2) we have the following particular cases:

(2)

p=R......... .(circum-circle); p = R secw.....(T. R. circle);

p=

R tano.....

...(cosine circle).

..........

Also area A'B'C' : ABC = sin2w: sin2(0+w) (Euc. VI. 19).

SECTION IV.

TUCKER'S CIRCLES, PARTICULAR CASES.

39. I. Cosine Circle. As a particular case of the general theorem (Art. 33 v.) we shall consider the antiparallels A'A", B'B", C'C" to pass through K. The points L, M, N will therefore coincide with K, which is also the centre of the corresponding Tucker's circle.

It is otherwise evident that the six segments KA', KA", etc., of antiparallels through K to the sides are equal. (Art. 31 (5)).

Also B'C'B"C", C'A'C"A", A'B'A"B" are rectangles since their diagonals are equal.

Again because A'B'B" is a right-angled triangle

A'B" B'B" cosA'B'B' = B'B" cosC,

=