From a point P on the basis BC of an isosceles draw parallels to the sides which intersect them at PB, PC. Show that the medial line L of PBPC passes through a fixed point (the circumcenter of the isosceles).
[1] Lines {L, PBPC, C'B'} are concurrent at a point Q on B'C', {B',C'} being the middles of the equal sides.
[2] Define point O as the intersection of L with the medial of B'C'. Triangles OC'PB and OB'PC are equal.
In fact, they have equal sides, since OC'=OB', OPB=OPC, O being on the two medial lines. C'PB=B'PC because of the isosceles: APC=BPB and AC'=B'P.
[3] It follows that quadrangles OQPCB' and OQC'PB are cyclic and their circumcircles have equal radii.
[4] It follows also that APBOPC is cyclic and OC', OB' are orthogonal to respective sides, thus O is the circumcenter of ABC.
Remarks
[1] Note that line L'' (not drawn), which is parallel to L from P, passes also from a fixed point A', which is the diametral of A on the circumcircle of ABC (A'P is parallel and double in length to OQ).
[2] One can carry out the construction of L and L' (orthogonal to L at Q) also for generic triangles. In that case line L' envelopes a parabola, known as the Artzt parabola of the triangle, which is tangent to sides AB, AC at B, C correspondingly.
[3] In the generic case L envelopes also a parabola, which for isosceli degenerates to a point as shown. For a discussion on these parabolas see the file Artzt_Generation.html .
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Isosceles intersection ![[0_0]](IsoscelesIntersection_files/IsoscelesIntersection_0_0_0.png)
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