See the attached image.

Let's apply the sine law to the triangles CAF and BAF:

`( C F ) / sin ( a + b ) = ( A F ) / sin ( pi / 2 - a + b ) , ` `( B F ) / sin ( a - b ) = ( A F ) / sin ( pi / 2 - a - b ) .`

Because `C F = B F , ` we obtain `sin ( a - b ) / sin ( a + b ) = cos ( a + b ) / cos ( a - b ) , ` or `sin ( 2a - 2b ) = sin ( 2a + 2b ) .`

This can mean two different things. First, it can mean that `2a - 2b = 2a + 2b , ` or `b = 0 , ` which is possible if and only if CAB is isosceles. In this case, the angle BAC may be any.

The second possibility is that `2a - 2b = pi - ( 2a + 2b ) , ` or `a = pi / 4 , ` so the angle BAC is a right angle.

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