Let I and J be the intersection of AD with EF and GB respectively. Using the sine rule on triangles
AFI: AI/sin(<AFI) = AF/sin(<AIF)
AGJ: AJ/sin(<AGJ) = AG/sin(<AJG)
AGF: AF/sin(<AGF) = AG/sin(<AFG)
If AD, EF, and GB are to be concurrent, then AI = AJ. Using this and the above three equations gives:
sin(<AIF) = sin(<AJG) or
sin(2 <BAC) = sin(180 - 4 <BAC) = sin(4 <BAC) = 2*sin(2 <BAC)*cos(2 <BAC) or
cos(2 <BAC) = 1/2 or
<BAC = 30
Note: If <BAC = 30, then the path segments are concurrent no matter what the value of <BAD
(although, the cue ball does not end up at vextex B).
Reflect triangle ABC about BC to get triangle A'BC.
Reflect triangle A'BC about A'C to get triangle A'B'C.
Reflect triangle A'B'C about A'B' to get triangle A'B'C'.
Reflect triangle A'B'C' about A'C' to get triangle A'B"C'.
The line AB" corresponds to the path taken by the cue ball.
It is clear from this figure that tan(<BAD) = sqrt(3)/5.
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