In Hyacinthos message #8782, you wrote:
>> Let ABC be a triangle and Pa a point.
>> The parallel to AB through Pa intersects BC at Aab
>> The parallel to AC through Pa intersects BC at Aac
>> Let A'ab, A'ac be the orthogonal projections of
>> Aab, Aac on AC, AB, resp.
>> Which is the locus of P such that AabA'ab = AacA'ac ?
This should be Pa.
Your condition is easily seen to be equivalent to the
condition that the B-parallelian and the C-parallelian
through Pa have equal length. (The B-parallelian of a
point P is the segment joining the points where the
parallel to CA through P meets AB and BC.)
Your locus is the line with trilinear equation
b(cz+ax) = c(ax+by).
>> The locus is a line La.
>> Similarly define the lines Lb, Lc.
>> The three lines La, Lb, Lc are concurrent.
They concur at the congruent parallelians point