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- Thread starter mirasjg
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- Thread starter
- #1

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- #2

- Feb 7, 2012

- 2,697

The picture is not there "as shown", but from your description it should look like this:

\begin{tikzpicture}

\draw (-3.5,1.3) circle (3.734cm) ;

\draw (2.1,1.3) circle (2.47cm) ;

\coordinate [label=right:$A$] (A) at (0,0) ;

\coordinate [label=right:$B$] (B) at (0,2.6) ;

\coordinate [label=above:$C$] (C) at (-4,5) ;

\coordinate [label=above:$D$] (D) at (-1,4.25) ;

\coordinate [label=above right:$M$] (M) at (0,4) ;

\coordinate [label=above:$E$] (E) at (1.5,3.75) ;

\coordinate [label=above:$F$] (F) at (4,3.1) ;

\draw (-4,5) -- (4,3) ;

\draw (0,-1) -- (0,5) ;

\end{tikzpicture}

As a hint, draw a circle centred at $M$, with radius $CM$. This circle passes through $C$ (obviously). It also passes through $F$ (almost as obviously) and through $A$ (why is that??). It then follows that the triangle $CMA$ is isosceles, so its base angles are equal. Use that fact to deduce that $BD$ is parallel to $AC$.