Let AOB, DOE be any two intersecting chords of a circle. Required to inscribe a circle in one of the compartments, as BOE.

Draw OF bisecting the angle BOE and let it cut the circle in F. Draw FM perpendicular to AB.

The idea of the following proof was communicated to me some years ago by Mr Edward Carpenter of Millthorpe, Derbyshire, formerly Fellow of Trinity Half, Cambridge; who remarked that it seemed to afford a demonstration of Taylor's Theorem which came very naturally and directly from the definition of a differential coefficient. The chief difficulty seemed to arise in dealing with the negligible small quantities which are produced in great numbers. However, I found it not difficult to complete the proof for the case when all the successive differential coefficients of f(x) are finite and continuous.

1. In the accompanying figure DEF is the pedal triangle of ABC and P, Q, R are the orthocentres of AFE, BDF, CED.

AP, BQ, CR evidently meet in the circumcentre, O, of ABC, which is the orthocentre of PQR,

2. Now the circumradius of AFE(ρa) = RcosA,

hence the sides of PQR are equal and parallel to the sides of DEF, i.e., the triangles are congruent, and their centre of perspective, L, bisects DP, EQ, FR.