Terris Becker edited untitled.tex  about 8 years ago

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&\iff (f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}d^{2}+\frac{1}{4}e^{2}\\  &\iff(f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}(d^{2}+e^{2})\\  \end{align}  The above equation is a circle with radius $r=\sqrt{\frac{1}{4}(d^{2}+e^{2})}$, so $r^{2}$ must be greater than or equal to zero in order for a solution to exist. Thus,  \begin{align}  \frac{1}{4}(d^{2}+e^{2}\geq0  \end{proof}