SOLUTIONOther points in the figure have been labelled as shown in Figure 1. Since a line tangent to a circle is perpendicular to the radius drawn to the point of tangency, then \(\Delta {\rm{ABC}}\) and \(\Delta {\rm{ABF}}\) are right triangles; and, clearly, \(\Delta {\rm{ADC}}\) and \(\Delta {\rm{AGF}}\) are also right triangles. \(\Delta {\rm{ABC}}\) and \(\Delta {\rm{ADC}}\) are a pair of similar right triangles. Thus, \(\angle {\rm{BAC}} = \angle {\rm{DAC}} = {\textstyle{1 \over 2}}\left( {2{\rm{\theta }}} \right) = {\rm{\theta }}\). It follows that \({\rm{CD}} = \tan {\rm{\theta }}\) and that \({\rm{CE}} = 1 - \tan {\rm{\theta }}\) (Figure 2). CE is...