13. If lines be drawn from the vertices of ABC to a point such that NBC=NCA-AB-0, prove that is given by the equation cot 0=cot A+ cot B+cot C.

[For sin30=sin(4-0) sin(B-0) sin(C-0); etc. Cf. Art. 28.] 14. In the general case if the lines in Ex. 13 making equal angles (a) with the sides are not concurrent, they form a triangle A'B'C' similar to ABC and the ratio of similitude is equal to

cos a-sin a(cot A+cot B+cot C): 1.

Defs. The Centres of Perspective of two lines AB and A'B' are the points of intersection of the pairs of lines AB', A'B and AA', BB' joining their extremities.

Two triangles are said to be in perspective when the lines joining corresponding vertices meet in a point. This point is called the Centre of Perspective of the triangles.

65. Criterion of Perspective of Triangles. Theorem. If the perpendiculars from the vertices of a triangle A'B'C' on the sides of another ABC be denoted by P1 P2 P3;

1'

91, 92, 93; T1, T2, T3 (i.e. from A′ on BCP1, A' on CAP2, and so on), the two are in perspective if

L12 P3-1; and conversely.

=

sin BAX'/sin CAX' =P3/P2

with similar values for r2/r1, and q1/93; multiplying these equations together, therefore, etc., by Art. 64, which also proves the converse* proposition.

66. Theorem.—If the vertices of two triangles connect concurrently, their pairs of corresponding sides intersect collinearly (BC and B'C' in X, etc. .....).

CX' P2 r1

Multiplying, we have

ᎪᏃ

C'Y

A'Y

BX CY AZ-T2 P3-1, therefore, etc.

Def. The line of collinearity is termed the Axis of Perspective or Homology† of the triangles.

EXAMPLES.

1. Any triangle escribed to a circle is in perspective with that formed by joining the points of contact of its sides.

[The centre of perspective is the symmedian point of the inscribed triangle.]

*Or thus :-Let O be the centre of perspective of the triangles and a, ß, y the perpendiculars from it on the sides of ABC; since ẞ/y=P2/P3, y/a = 93/91, and a/ß=r1/r2; multiply and reduce; therefore, etc.

†The term Homology is due to Poncelet who first studied the properties of homological figures in space, v. Traité des propriétés projectives des figures (1822).

2. If three triangles ABC, A,B,C1 A,B,C2 have a common axis of perspective XYZ, their centres of perspective when taken two and two are collinear.

2

[For the triangles (fig. of Ex. 3) BBB, and CCC, are in perspective, their centre being at X; similarly Y is the centre of perspective of CCC, AA14, and Z of AAA, and BB,B2. Hence the corresponding sides of these pairs of triangles intersect in collinear points. But these points (e.g. AA1, BB1) are the centres of perspective of the given triangles in pairs; therefore, etc.]

3. If three triangles ABC, A,B,C1, A,B,C2 have a common centre of perspective, their axes are concurrent.

[Consider the three triangles whose sides are respectively the directions BC, B1C1, B2C2; CA, C141, C2A2; AB, A1B1, A2B2

It is manifest they are in pairs in perspective, the axis of the first pair being CC1; and XY is a line joining corresponding vertices.

Thus the axis of perspective XY of any two and therefore of every two of the given triangles passes through the centre of perspective of the conjugate triad ]

NOTE.--It will be noticed that the common centre 0 of the three given triangles is the point of concurrence of the axes AA1, BB1 CC, of the conjugate triad, and the common centre of the conjugate triad taken in pairs is the point of concurrence of the axes of the given triangles.

4. Brocard's first triangle is in perspective in three ways with ABC.

[The Brocard points are evidently two centres of perspective (Art. 28); also the lines AA', BB', CC' are concurrent, for p2/p3 found by aid of the property of Art. 28, Ex. 2, to be c3/b3; therefore, etc. The three centres of perspective are the mean centres of the vertices ABC for multiples proportional to (Art. 52)

1, 1, 1 1. 1. 1 1. 1. 1

,

5. If N,N',N" denote the three centres of perspective of ABC and its first Brocard triangle A'B'C', to prove that the corresponding vertices of their three median triangles lie on three right lines. (Stoll.)

[For A'B'C' and ABC have a common centroid G (Art. 53, Ex. 6). But '" has the same centroid; for its vertices are the mean 1 1 1 1 1 1

α

centres of A, B, C for multiples proportional to b2c2a2' c2 a2 b2; therefore (Art. 53, Cor. 3) the mean centre of , N', N" is

1. 1. 1

2

N,

that for A, B, C for multiples each =1/a2+1/b2+1/c2. Now let L, L, L" be the middle points of the corresponding sides of the three triangles such that GA=2GL, GA'=2GL, and GO"=2GL": since A, A', " are collinear; L, L, L" are also collinear, and the two lines of collinearity parallel.]

67. Theorem.-Two triangles ABC and A'B'C' are in

where X and X' are the points of intersection of BC with C'A' and A'B', etc.; and conversely.

Using the previous notation, we have by similar

COR. 1. Pascal's Theorem.-If XX'YY'ZZ' be any cyclic hexagon, then (Euc. III. 36)

AY.AY' AZ. AZ'; BZ. BZ' =BX . BX',

=

etc.

Hence :-The two triangles formed by the two triads of alternate sides of any cyclic hexagon are in perspective; or, the opposite sides of a cyclic hexagon meet in three collinear points.

The centre and axis of perspective of any two triangles in perspective are called the Pascal Point and Line of the hexagon XX'YY'ZZ', which is termed a Pascal Hexagon.

COR. 2. If X, X'; Y, Y'; Z, Z' coincide in pairs on the circle, the sides of the hexagon become the tangents to the circle at X, Y, Z, and the chords of contact YZ, ZX and XY; the Pascal point is therefore the symmedian point of the triangle XYZ. (Art. 66, Ex. 1.)

* When only sixteen years old, Pascal discovered this property of the mystic hexagram. Essai sur les Coniques, Pascal, 1640.