Van Schooten offered a second solution to the problem of bisecting an angle BAC. He wrote:

Another way: Assuming the point B is in AB, find in CA, produced through A, the segments AD, DE, each equal to AB, and draw the straight line through the points E, B, and put on it BF equal to BE, and join AF. I say it bisects that angle BAC.

The proof of correctness of this construction may not be immediately obvious. As a hint, we suggest that a line from B to D might be useful.