goodyear2013 wrote:
Is quadrilateral ABCD a square?
(1) AB=BC
(2) ABCD is a rectangle.
\(ABCD\,\,\mathop = \limits^? \,\,{\text{square}}\)
\(\left( 1 \right)\,\,AB = BC\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{trivial}}\,\,{\text{geometric}}\,\,{\text{bifurcation}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.\)
\(\left( 2 \right)\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,{\text{ABCD}}\,\,{\text{rectangle}}\,\,{\text{non - square}} \hfill \\
\,{\text{Take}}\,\,{\text{ABCD}}\,\,\left( {{\text{rectangle}}\,\,{\text{and}}} \right)\,\,{\text{square}} \hfill \\
\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{INSUFF}}.\)
\(\left( {1 + 2} \right)\,\,\,\left\{ {\,\left. \begin{gathered}
\,{\text{rectangle}}\,\,\,\, \Rightarrow \,\,\,\,{\text{parallelogram}} \hfill \\
\,\,\left[ {AB = BC} \right]\,\, \cap \,\,{\text{parallelogram}}\,\,\,\, \Rightarrow \,\,\,\,{\text{rhombus}} \hfill \\
\end{gathered} \right\}} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{\text{SUFF}}{\text{.}}\)
\(\left( * \right)\,\,\,\left\{ \begin{gathered}
\,{\text{rectangle}} \hfill \\
\,{\text{rhombus}} \hfill \\
\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{square}}\)
This solution follows the notations and rationale (
quadrilaterals properties) taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik ::
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