(Nulty's Geom. p. ). But the point E, since the angle ADE is a right one, will likewise fall in the circumference of a semicircle described upon the diameter AD (Nulty's Geom. P. ). And therefore FE, being a radius, must be equal to AF; and consequently DG AD, supposing DC produced to meet AQ in G.

=

Therefore, in order to the geometrical construction, having made Ab perpendicular and equal to AB, and drawn AD to the middle of Cb (as above intimated), let DG, in DC produced, be taken equal to AD; and from G, through A, draw GP, and the thing is done.

It often happens that the demonstration of a geometrical construction, to be the most neat and elegant, proceeds upon principles very different from those whereby we first arrived at such construction. The case above is an instance of it; where, from the similar triangles, it is manifest that GQ (Ap): QC :: Gp (AQ): pb; and therefore AQ × QC = {Ap X pb=1BP X AP.

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