# P.o.t.W. #4 Solution 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...

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