Question 171
In a triangle $ABC$ with an angle $\angle A = 60^\circ$, the Euler line intersects the circumcircle of the triangle at the points $U$ and $V$. $AD$ is the height of $\Delta ABC$ and $\omega$ is the nine-point circle. The circumcircle $\Omega$ of triangle $\Delta UVD$ intersects the $\omega$ for the second time at the point $K$. $S$ is a point such that $KS$ is a diameter of $\omega$. Prove that $AS$ is the symmedian of $\Delta ABC$.
*An Euler line is a line through the orthocenter, the circumcenter, and the point where the medians of the triangle intersect.*
*The nine-point circle is the circle on which the feet of the heights, the midpoints of the sides, and the midpoints of the segments between the vertices and the orthocenter lie.*
*The symmedian of a triangle from a vertex is a line symmetric to the median from the vertex about the bisector from the vertex.*
Figure 23
*An Euler line is a line through the orthocenter, the circumcenter, and the point where the medians of the triangle intersect.*
*The nine-point circle is the circle on which the feet of the heights, the midpoints of the sides, and the midpoints of the segments between the vertices and the orthocenter lie.*
*The symmedian of a triangle from a vertex is a line symmetric to the median from the vertex about the bisector from the vertex.*
Figure 23