Symmedian

Many times in geometry, if we take three special lines through the vertices of a triangle, or cevians, then their reflections about the corresponding angle bisectors, called isogonal lines, will also have interesting properties. For instance, if three cevians of a triangle intersect at a point P, then their isogonal lines also intersect at a point, called the isogonal conjugate of P.

The symmedians illustrate this fact.

• In the diagram, the medians (in black) intersect at the centroid G.
• Because the symmedians (in red) are isogonal to the medians, the symmedians also intersect at a single point, L.

This point is called the triangle's symmedian point, or alternatively the Lemoine point or Grebe point.

The dotted lines are the angle bisectors; the symmedians and medians are symmetric about the angle bisectors (hence the name "symmedian.")

AD is the symmedian through vertex A of △ABC.

Let △ABC be a triangle. Construct a point D by intersecting the tangents from B and C to the circumcircle. Then AD is the symmedian of △ABC.[2]

first proof. Let the reflection of AD across the angle bisector of ∠BAC meet BC at M'. Then:

${\displaystyle {\frac {|BM'|}{|M'C|}}={\frac {|AM'|{\frac {\sin \angle {BAM'}}{\sin \angle {ABM'}}}}{|AM'|{\frac {\sin \angle {CAM'}}{\sin \angle {ACM'}}}}}={\frac {\sin \angle {BAM'}}{\sin \angle {ACD}}}{\frac {\sin \angle {ABD}}{\sin \angle {CAM'}}}={\frac {\sin \angle {CAD}}{\sin \angle {ACD}}}{\frac {\sin \angle {ABD}}{\sin \angle {BAD}}}={\frac {|CD|}{|AD|}}{\frac {|AD|}{|BD|}}=1}$

second proof. Define D' as the isogonal conjugate of D. It is easy to see that the reflection of CD about the bisector is the line through C parallel to AB. The same is true for BD, and so, ABD'C is a parallelogram. AD' is clearly the median, because a parallelogram's diagonals bisect each other, and AD is its reflection about the bisector.

third proof. Let ω be the circle with center D passing through B and C, and let O be the circumcenter of △ABC. Say lines AB, AC intersect ω at P, Q, respectively. Since ∠ABC = ∠AQP, triangles △ABC and △AQP are similar. Since

${\displaystyle \angle PBQ=\angle BQC+\angle BAC={\frac {\angle BDC+\angle BOC}{2}}=90^{\circ },}$

we see that PQ is a diameter of ω and hence passes through D. Let M be the midpoint of BC. Since D is the midpoint of PQ, the similarity implies that ∠BAM = ∠QAD, from which the result follows.

fourth proof. Let S be the midpoint of the arc BC. |BS| = |SC|, so AS is the angle bisector of ∠BAC. Let M be the midpoint of BC, and It follows that D is the Inverse of M with respect to the circumcircle. From that, we know that the circumcircle is an Apollonian circle with foci M, D. So AS is the bisector of angle ∠DAM, and we have achieved our wanted result.

The concept of a symmedian point extends to (irregular) tetrahedra. Given a tetrahedron ABCD two planes P, Q through AB are isogonal conjugates if they form equal angles with the planes ABC and ABD. Let M be the midpoint of the side CD. The plane containing the side AB that is isogonal to the plane ABM is called a symmedian plane of the tetrahedron. The symmedian planes can be shown to intersect at a point, the symmedian point. This is also the point that minimizes the squared distance from the faces of the tetrahedron.[3]

• Symmedian and Antiparallel at cut-the-knot
• Symmedian and 2 Antiparallels at cut-the-knot
• Symmedian and the Tangents at cut-the-knot
• An interactive Java applet for the symmedian point
• Isogons and Isogonic Symmetry