The following proof of the identity

Where A, B, C are the angles of a triangle differs from that given by Dr Bell in the last number of Mathematical Notes and from that given in “New Trigonometry for Schools” by Lock and Child in that the area of the triangle is made use of.

Let I be the incentre of ΔABC, and let the excentre opposite A be I1 Draw perpendiculars IF and I1F1 to AB. IBI1 = 90°.

If P (x′, y′) is the given point and ax + by + c = 0 the equation of the given line, the expression for the distance of P from the line and the coordinates of the foot of the perpendicular from P to the line can be obtained by projections as follows:

Let NP = R and have projections X and Y on OX and OY respectively. Then if NP makes an angle θ with OX

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For determining the Cartesian co-ordinates of the centre of curvature of a plane curve two methods are principally used in the text-books. One of these (see for example, Edwards, “Differential Calculus,” p 266, § 339) having previously established the formula for the radius of curvature, derives the coordinates of the centre of curvature by using the circular functions of the angle “ψ,” which the tangent to the curve at the point considered makes with OX. Since, however, for the same tangent, and therefore the same “ψ” the curve may be either convex or concave towards OX (and accordingly the centre of curvature have quite different positions), this method introduces an essential ambiguity, and is not to be recommended.