Some of the ISLs I did before INMO :P [2005 G3]: Let ABCD be a parallelogram. A variable line g through the vertex A intersects the rays BC and DC at the points X and Y, respectively. Let K and L be the A-excenters of the triangles ABX and ADY. Show that the angle \measuredangle KCL is independent of the line g Solution: Note that \Delta LDK \sim \Delta XBK
and \Delta ADY\sim \Delta XCY.
So we have \frac{BK}{DY}=\frac{XK}{LY}
and \frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.
Hence \frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY}
\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}
\implies \Delta CBK\sim \Delta LDK
And we are done. We get that \angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2
Motivation: I took a hint on this. I had other angles but I did...