58-Two nice line in a triangle

The first line: 

Let  be a triangle, a line  meets  at  are midpoints of .  are reflection of  in . Then  are collinear.

Solution:

http://www.cut-the-knot.org/pythagoras/AMatterOfCollinearity.shtml

The Secon line:

Let  be a triangle,  are projection of  on .  be a point on the plane.  lie on  respectively, define  cyclically such that .  is intersection of perpendicular of  at . Define  cyclically. Then  are collinear. Please see the figure attachments

Solution: by Telv Cohl https://www.facebook.com/telv.cohl?fref=ts

Let A4 B4 C4 be the mid point of GA3 GB3 GC3 and H' be the orthocenter of triangle A1B1C1. Easy to see GA3 is the diameter of circle(GA1A2A3)
and A4 is the center of circle(GA1A2A3) similarity for B3 B4 and C3 C4

Since A1H' x H'A2=B1H' x H'B2=C1H' x H'C2 so (GA1A2A3) (GB1B2B3) (GC1C2C3) are coxial ( with common radical axis GH' ) hence A4 B4 C4 are collinear ie.A3 B3 C3 are collinear