In geometry, symmedians are three particular lines associated with every triangle. They are constructed by taking a median of the triangle (a line connecting a vertex with the midpoint of the opposite side), and reflecting the line over the corresponding angle bisector (the line through the same vertex that divides the angle there in half). The angle formed by the symmedian and the angle bisector has the same measure as the angle between the median and the angle bisector, but it is on the other side of the angle bisector. The three symmedians meet at a triangle center called the Lemoine point. Ross Honsberger has called its existence "one of the crown jewels of modern geometry". Isogonality 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 , then their isogonal lines also intersect at a point, called the isogonal conjugate of . The symmedians illustrate this fact. In the diagram, the medians (in black) intersect at the centroid . Because the symmedians (in red) are isogonal to the medians, the symmedians also intersect at a single point, . 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.") Construction of the symmedian Let be a triangle. Construct a point by intersecting the tangents from and to the circumcircle. Then is the symmedian of . first proof. Let the reflection of across the angle bisector of meet at . Then: second proof. Define as the isogonal conjugate of . It is easy to see that the reflection of about the bisector is the line through parallel to . The same is true for , and so, is a parallelogram. is clearly the median, because a parallelogram's diagonals bisect each other, and is its reflection about the bisector. third proof. Let be the circle with center passing through and , and let be the circumcenter of . Say lines intersect at , respectively. Since , triangles and are similar. Since we see that is a diameter of and hence passes through . Let be the midpoint of .