will be seen that if L. H. rotation take place the point P must move along a straight line T U parallel to B C, (for P being the centre of the curve R Q 8 is equidistant from all points in it), and similarly the point Q moves in a straight line Q T parallel to A B. The two paths intersect at T at an angle of 60 the same angle, namely, as that at which P S and Q S intersect in S. The as yet unknown path of the point S relatively to the figure ABC is therefore simply that of the vertex of a triangle PS ft of which the ends P and Q of the base slide upon the arms of an angle equal to the vertex angle of the triangle itself. Let P Q S,
rs
Fig. 94.
Fig. 94, be such a triangle, a, and 7 the angles at its vertex and base respectively. UTQ = a, is the angle upon the arms of which the points P and Q slide. The points $ and T are, however, points in a circle passing through P and Q, a being the angle at the circumference upon the chord PQ. If, then, we join 8 and T, we have Z. Q T S equal to the angle /3 at P, the circum- ferential angle upon the chord QS, and therefore constant for all positions of the triangle. If, further, ST be produced through Tto A, the angle A TP = 180 -(a+^); that is, = the base angle 7 at Q. The point S therefore moves in a straight line, which makes with the arms of the given angle angles respectively equal to those at the base of the given triangle. This straight line is in Fig. 93 above the third side A C of the triangle, which makes at T with U T and Q T the angle GO the base angle of the equiangular (and also equilateral) triangle PSQ; all three sides of the triangle therefore touch the duangle continuously.
