Dear Antreas,

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

X(192).

Sincerely,

Darij Grinberg