The base AB of a right-angled triangle ABC is divided into n equal parts. On these parts as bases n outer rectangles (Fig. 1) and(n–1) inner rectangles (Fig. 2) are drawn. To find Snthe sum of the areas of the outer set and Sn′ the sum of the areas of the inner set.

The equation of the polar of a point (x′, y′) with respect to a circle x2 + y2 = r2 can be found very easily by the use of

either the perpendicular or the intercept form of the equation of a straight line.

Theorem I. If to each of the elements of any row of a determinant there be added the same fraction of the difference between the corresponding elements in two other rows, the value of the determinant is unaltered. This is a case of a well known elementary theorem.

An application of the method of Induction to the proof of a general result or theorem should involve two steps: (i) the discovery of the result to be proved by the consideration of particular cases; (ii) a proof, by Mathematical Induction, or otherwise, that the result, found to be true in particular cases, is universally true.

It is evident that if MCA1 = MBC the point M lies on the circle O, which touches AC at C and passes through B. Again, if we take two equal and opposite pencils A(BIMC) and C(A1EMb) we know that their intersections lie on an hyperbola, and observe that when AI is the bisector of A its corresponding ray CE of the second pencil is parallel to it. This gives the direction of the asymptotes; the second asymptote will be parallel to the external bisector of A. The conic is therefore a rectangular hyperbola. When the ray of the first pencil coincides with AC1 the ray of the second corresponding to it will make with CA an angle equal to A, it is therefore parallel to AB, and the point of intersection is C; hence the second ray ab is a tangent at C, similarly BA is a tangent at A; therefore AC passes through the centre of the conic, and B′ is the centre. The point M where this conic cuts O is the required point M. Denote this hyperbola by γ, then on repeating this construction at B we get a second hyperbola β cutting the circle O1 at M1, which makes angle M1BA2 = M1AC = M1CB. (See Fig. on p. 157.)