<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

\begin{description}
\item[(a)] The midpoints of consecutive sides of an arbitrary
quadrilateral are joined.  Show that the figure so formed is a
parallelogram.
\item[(b)] Show that there is a triangle with sides equal and
parallel to the medians of any given triangle.
\item[(c)] The points $D,E,F$ divide the sides $AB, BC, CA,$
respectively, of the triangle $ABC$ in the ration $m:n$.  Show
that for any point $P$ in space $$\vec{PA} + \vec{PB} + \vec{PC} =
\vec{PD} + \vec{PE} + \vec{PF}$$
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
${}$

\setlength{\unitlength}{.5in}
\begin{picture}(4,4)
\put(1,0){\line(-1,2){1}}

\put(0,2){\line(1,1){2}}

\put(2,4){\line(1,-1){2}}

\put(4,2){\line(-3,-2){3}}

\put(0.5,1){\line(1,4){0.5}}

\put(1,3){\line(1,0){2}}

\put(3,3){\line(-1,-4){0.5}}

\put(2.5,1){\line(-1,0){2}}

\put(0,2.3){\makebox(0,0){A}}

\put(2,4.3){\makebox(0,0){B}}

\put(4.3,2){\makebox(0,0){C}}

\put(1.3,0){\makebox(0,0){D}}

\put(0.8,3){\makebox(0,0){P}}

\put(3.3,3.1){\makebox(0,0){Q}}

\put(2.5,0.7){\makebox(0,0){R}}

\put(0.3,1){\makebox(0,0){S}}

\end{picture}


Choose an origin $O$ then:

$\begin{array}{rcl} \vec{OP} &amp; = &amp; \frac{1}{2}(\vec{OA} +
\vec{OB}) \\ \vec{OQ} &amp; = &amp; \frac{1}{2}(\vec{OB} + \vec{OC})  \\
\vec{OR} &amp; = &amp; \frac{1}{2}(\vec{OC} + \vec{OD}) \\ \vec{OS} &amp; = &amp;
\frac{1}{2}(\vec{OD} + \vec{OA}) \\ \vec{PQ}  = \vec{OQ} -
\vec{OP} &amp; = &amp; \frac{1}{2}(\vec{OC} - \vec{OA}) \\ \vec{SR}  =
\vec{OR} - \vec{OS} &amp; = &amp; \frac{1}{2}(\vec{OC} - \vec{OA})
\end{array}$

SO $PQ = RS$ and $PQ$ is parallel to $RS$.  So $PQRS$ is a
parallelogram.

${}$
\newpage
\item[(b)]

${ }$

\setlength{\unitlength}{.5in}
\begin{picture}(4,3)
\put(0,.5){\line(1,1){2}}

\put(0,0.5){\line(1,0){3}}

\put(2,2.5){\line(1,-2){1}}

\put(1,1.5){\line(2,-1){2}}

\put(1.5,0.5){\line(1,4){0.5}}

\put(0,0.5){\line(5,2){2.5}}

\put(0,0.3){\makebox(0,0){$A$}}

\put(2,2.8){\makebox(0,0){$B$}}

\put(3.2,0.5){\makebox(0,0){$C$}}

\put(.9,1.6){\makebox(0,0){$P$}}

\put(2.75,1.6){\makebox(0,0){$Q$}}

\put(1.5,.2){\makebox(0,0){$R$}}

\put(4,0.5){\line(0,1){2}}

\multiput(1.5,0.5)(.25,.1){11}{\circle*{.05}}

\multiput(2,2.5)(.2,-.1){11}{\circle*{0.05}}

\put(4.3,1.5){\makebox(0,0){$T$}}

\end{picture}

Choose $B$ to be the origin, and let $\vec{BA} = {\bf a}
\hspace{.2in} \vec{BC} = {\bf c}$.

Then $\vec{BP} = \frac{1}{2} {\bf a} \hspace{.2in} \vec{BQ} =
\frac{1}{2}{\bf c} \hspace{.2in} \vec{BR} = \frac{1}{2}({\bf
a}+{\bf c})$

$\begin{array}{rcl} \vec{AQ}  = \vec{BQ} - \vec{BA} &amp; = &amp;
\frac{1}{2}({\bf c} - {\bf a}) \\ \vec{CP}  = \vec{BP} - \vec{BC}
&amp; = &amp; \frac{1}{2}({\bf a} - {\bf c}) \end{array}$



$\begin{array}{rcl} {\rm Let\ the\ point\ }T{\rm \ be\ defined\
by\ }\vec{BT} &amp; = &amp; \frac{1}{2}(\vec {BR} + \vec {AQ})\\  &amp; = &amp;
\frac{1}{2}({\bf a} + {\bf c}) + \frac{1}{2} {\bf c} - {\bf a} \\
&amp; = &amp; {\bf c} - \frac{1}{2} {\bf a}\end{array}$

Let the point $R$ be defined by $\vec{BR} = \vec{PC} = {\bf c} -
\frac{1}{2}{\bf a}$

So $\vec{BR} = \vec{BT}$ and thus $R=T$

So the triangle $BRT$ is as required, since $\vec{RT} = \vec{BT} -
\vec{BR} = \vec{AQ}$ and $\vec{BT} = \vec{PC}.$

Note if the medians form a triangle the sum of the vectors
represented then should be zer0, and it is.

${}$

\item[(c)]
By the ration theorem: \begin{eqnarray*} \vec{PD} &amp; = &amp;
\frac{m\vec{PB} + n\vec{PA}}{m+n} \\ \vec{PE} &amp; = &amp;
\frac{m\vec{PC} + n\vec{PB}}{m+n} \\ \vec{PF} &amp; = &amp;
\frac{m\vec{PA} + n\vec{PC}}{m+n} \end{eqnarray*} Adding gives
$\vec{PD} + \vec{PE} + \vec{PF} = \vec{PA} + \vec{PB} + \vec{PC}$










\end{description}
\end{document}
</pre></body></html>