|Theorem : PQR are in line. The pole of PQR is the Lemoine point of ABC.|
A few words about what we'll prove here. First of all about the Lemoine point of a triangle.
We call "symmedian" the reflection of a median on the angle bissector.
The 3 symmedians concur, the intersection point is named the Lemoine point of the triangle.
This is a direct consequence of the angle version of the Ceva theorem, and from the medians being concurrent.
Points P, Q and R in line is a specific case
of Pascal theorem about conic sections :
Let 6 points on a conic section ABCDEF.
The intersect points of AD with BF, of AE with CF, and of BE with CD are in line.
When D→A, E→B et F→C and when the conic section is a circle, this gives the present property.
Here we'll prove P,Q,R in line directly, without the use of Pascal theorem.
This will also give the pole of line PQR.
Let U, V, W the intersect points of tangents in A, B and C.
By construction U is the pole of BC, hence U is conjugate of all points on BC, specifically of P.
All points of the tangent in A are conjugate of A, hence P is conjugate of A.
The polar of P is then line UA.
Similarily polar of Q is line VB and polar of R is line WC.
These three lines intersect in L which is the Gergonne point of triangle UVW.
Just because the circumcircle of ABC is the incircle of UVW and A,B,C are the contact points.
Recall : The proof that UA, VB and WC intersect in one point is immediate from the Ceva theorem.
UB = UC, VC = VA et WA = WB, hence AV/AW × CV/CU × BU/BW = -1.
Point L is then conjugate of P, Q and R. Hence P, Q and R are on the polar of L, hence are in line.
We have now to prove that lines AU, BV and CW are the symmedians of ABC.
This will prove the required property (pole of PQR is the Lemoine point of ABC) and also will prove that in any triangle (UVW), the Gergonne point is the Lemoine point of the inscribed triangle (ABC), build from the contact points of incircle.
Let O the circumcenter of ABC, and M the intersect point of UO with BC. M is midpoint of BC.
UA intersects the circumcircle in E and intersects BC in D. BC being the polar of U, UEDA is an harmonic range.
Draw the parallels to BC in U, E, D(!) and A. They intersect any secant into an harmonic range.
The parallel in A intersects UO in X and intersects the circumcircle in A'.
The parallel in E intersects UO in Y and intersects the circle in E'.
UYMX is then an harmonic range.
UO is a symmetry axis hence A'E'U are in line.
M is then the intersect point of AE' and A'E. (classical construction of conjugate M of U).
Angles BAE and E'A'C are equal from symmetry, and inscribed angles E'AC and E'A'C are equal.
Hence angles BAU and MAC are equal, and as AM is the median, AU is the symmedian. QED.
The ABC circumcircle is incircle of UVW only if ABC is acute (all angles acute)
If ABC has an obtuse angle, let A, the circumcircle is an excircle in angle U of UVW.
All the above properties are still valid, but point L is no more the Gergonne point of UVW.
It is still the Lemoine point of ABC and the pole of PQR.
This doesn't matter to prove that the Gergonne point is the Lemoine point of the inscribed triangle, as the inscribed triangle is always acute.
The proof that PQR are in line is as already said a direct result of Pascal theorem,
which is about any conic section.
"The" conic section circumscribed to ABC has no meaning (there are infinitely many), so we may change the statement into :
Let any conic section and three points ABC on this one...
Lines AU, BV, CW are still concurrent from the dual of Pascal theorem :
the Brianchon theorem.
Point L is still the pole of PQR. But it is not related with any Gergonne point, nor with symmedians or Lemoine point...
points P, Q, R are the centers of Apollonius circles of ABC
Line CC', radical line of the two circles, is then also the polar of R from the circumcircle.
Let's proof that this polar is a symmedian.
Let M3 the midpoint of AB. The perpendicular bissector MM3 of AB
intersects the circumcircle in C3, midpoint of arc AB opposite to C.
Hence CC3 is angle bissector of angle C. CM3 is the median.
Draw circle with diameter RM. It goes at C and C' (CM_|_CR) and M3 (MM3_|_M3A).
The inscribed angles in that circle give : C'MM3 = C'CM3
Also in the circumcircle, angle C'MC3 is double of inscribed angle C'CC3
Hence C'CM3 = 2 * C'CC3, that is CC3 is angle bissector of angle C'CM3, that is line CC' is reflection of median CM3 on angle bissector of angle C : CC3. Hence :
The polar of R is a symmedian
Finally, as the symmedians concur in Lemoine point, their poles (points P,Q,R) are on a same line, the polar of L. QED.