MHB Can You Prove that AC//BD and AF//BE are Parallel in this Geometry Problem?

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In the geometry problem involving two intersecting circles, a straight line CDEF is formed by extending chord CD to points E and F on the second circle. The common chord AB intersects line CF at point M, which is the midpoint of CF, and angle CAF is established as 90 degrees. To prove that lines AC and BD are parallel, one can construct a circle centered at M with radius CM, which also passes through points C, F, and A. This configuration leads to triangle CMA being isosceles, indicating that its base angles are equal, thus establishing that BD is parallel to AC. The conclusion is that both AC//BD and AF//BE are confirmed as parallel lines.
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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.
 
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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 $AC$.
 
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