## Re: Altitudes bisected etc

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• Dear Antreas, ... ^ 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
Message 1 of 3 , Dec 7, 2003
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
• Dear Antreas, ... No, since your equation entails B C || BC, C A || CA and A B || AB, so that ABC is the antimedial triangle of A B C , and P is the
Message 2 of 3 , Dec 7, 2003
Dear Antreas,

In Hyacinthos message #8781, you wrote:

>> Let ABC be a triangle and AA', BB', CC' the cevian
^^^^^^^^^^^^^
>> triangle of P.
>>
>> For which points P we have that:
>>
>> AP / PA' = BP / PB' = CP / PC' ?
>>
>> Are there other than the centroid points with that
>> property?

No, since your equation entails B'C' || BC, C'A' || CA
and A'B' || AB, so that ABC is the antimedial triangle
of A'B'C', and P is the centroid of triangle ABC.

Sincerely,
Darij Grinberg
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