proof of parallelogram theoremsProofOfParallelogramTheorems2007-06-14 18:53:262007-06-16 09:58:08Proofparallelogram theorems0% this is the default PlanetMath preamble. as your knowledge
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\begin{thm} \label{thm:one}The opposite sides of a parallelogram are congruent.
\end{thm}
\begin{proof} This was proved in the \PMlinkname{parent}{ParallelogramTheorems} article.
\end{proof}
\begin{thm} \label{thm:two}If both pairs of opposite sides of a quadrilateral are congruent, the quadrilateral is a parallelogram.
\end{thm}
\begin{proof}
Suppose $ABCD$ is the given parallelogram, and draw $\ol{AC}$.
\begin{center}\includegraphics{Parallelogram2}\end{center}
Then $\triangle ABC\cong\triangle ADC$ by SSS, since by assumption $AB=CD$ and $AD=BC$, and the two triangles share a third side.
By CPCTC, it follows that $\angle BAC\cong\angle DCA$ and that $\angle BCA\cong \angle DAC$. But the theorems about corresponding angles in transversal cutting then imply that $\ol{AB}$ and $\ol{CD}$ are parallel, and that $\ol{AD}$ and $\ol{BC}$ are parallel. Thus $ABCD$ is a parallelogram.
\end{proof}
\begin{thm} \label{thm:three}If one pair of opposite sides of a quadrilateral are both parallel and congruent, the quadrilateral is a parallelogram.
\end{thm}
\begin{proof}
Again let $ABCD$ be the given parallelogram. Assume $AB=CD$ and that $\ol{AB}$ and $\ol{CD}$ are parallel, and draw $\ol{AC}$.
\begin{center}\includegraphics{Parallelogram3}\end{center}
Since $\ol{AB}$ and $\ol{CD}$ are parallel, it follows that the alternate interior angles are equal: $\angle BAC\cong \angle DCA$. Then by SAS, $\triangle ABC\cong \triangle ADC$ since they share a side.
Again by CPCTC we have that $BC=AD$, so both pairs of sides of the quadrilateral are congruent, so by Theorem \ref{thm:two}, the quadrilateral is a parallelogram.
\end{proof}
\begin{thm} \label{thm:four}The diagonals of a parallelogram bisect each other.
\end{thm}
\begin{proof}
Let $ABCD$ be the given parallelogram, and draw the diagonals $\ol{AC}$ and $\ol{BD}$, intersecting at $E$.
\begin{center}\includegraphics{Parallelogram4}\end{center}
Since $ABCD$ is a parallelogram, we have that $AB=CD$. In addition, $\ol{AB}$ and $\ol{CD}$ are parallel, so the alternate interior angles are equal: $\angle ABD\cong \angle BDC$ and $\angle BAC\cong \angle ACD$. Then by ASA, $\triangle ABE\cong \triangle CDE$.
By CPCTC we see that $AE=CE$ and $BE=DE$, proving the theorem.
\end{proof}
\begin{thm} \label{thm:five}If the diagonals of a quadrilateral bisect each other, the quadrilateral is a parallelogram.
\end{thm}
\begin{proof}
Let $ABCD$ be the given quadrilateral, and let its diagonals intersect in $E$.
\begin{center}\includegraphics{Parallelogram5}\end{center}
Then by assumption, $AE=EC$ and $DE=EB$. But also vertical angles are equal, so $\angle AED\cong \angle AEB$ and $\angle CED\cong \angle AEB$. Thus, by SAS we have that $\triangle AED\cong \triangle CEB$ and $\triangle CED\cong \triangle AEB$.
By CPCTC it follows that $AB=CD$ and that $AD=BC$. By Theorem \ref{thm:one}, $ABCD$ is a parallelogram.
\end{proof}