rectangle Rectangle 2001-12-12 00:04:08 2004-03-08 17:47:30 Definition \usepackage{graphicx} %\usepackage{xypic} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand{\mathbb}[1]{\mathbbmss{#1}} \newcommand{\figura}[1]{\begin{center}\includegraphics{#1}\end{center}} \newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}} \newtheorem{dfn}{Definition} \textbf{Rectangle}. A quadrilateral whose four angles are equal, that is, whose 4 angles are equal to $90^\circ$. \figura{rectangulo} \emph{\bf Any rectangle is a parallelogram.}\\ This follows from angles $\angle BAD $ and $\angle ADC$ adding up $180^\circ$. Since parallelograms have their opposite sides equal, so do rectangles. In the picture, $AB=CD$ and $BC=DA$. Rectangles are the only parallelograms to be also cyclic (since opposite angles add up $180^\circ$. Notice that every square is also a rectangle, but there are rectangles that are not squares Rectangles have their two diagonals equal (since triangles $ABC$ and $ABD$ are congruent), A nice result following from this is that the quadrilateral obtained by joining the midpoints of the sides is a rhombus. \figura{rectangulo2} Since $PQ$ joins midpoints of sides in triangle $ABC$, we have $PQ=CA/2$. Similarly we have $RS=CA/2$, $QR=BD/2$ and $SP=BD/2$ and thus the sides of quadrilateral $PQRS$ are all equal, in other words, $PQRS$ is a rhombus.