mirasjg said:
Two circles meet at A and B. A chord CD of one circle is produced to meet the other circle at E and F so that CDEF is a straight line, as shown. The common chord AB is produced to meet the line CF at a point M between D and E. If M is the midpoint of CF and angle CAF=90 degrees, prove that AC//BD and AF//BE are parallel.
The picture is not there "as shown", but from your description it should look like this:
[TIKZ]\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) ;
[/TIKZ]
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 $AC$.