62-Similar Goormaghtigh theorem, with Incenter

Problem: $I$ is incenter(or excenter), $A',B',C'$ lie on $IA,IB,IC$ such that $\frac{IA}{IA'}=\frac{IB}{IB'}=\frac{IC}{IC'}$ then three perpendicular of $IA,IB,IC$ at $A',B',C$ meets three sidelines at $A_1,B_1,C_1$ then $A_1,B_1,C_1$ are collinear.

Solution by: Telv Cohl https://www.facebook.com/telv.cohl

Let incircle of triangle $ABC$ touch $BC, CA, AB$ at $D, E, F$ and denote circle with diameter $RQ$ as circle $(RQ)$ 

Inverse with center $I$ and power $ID^2=IE^2=IF^2$ this inversion map $A, B, C, A', B', C'$ into $X, Y, Z ,X', Y', Z'$ and map $A_1, B_1 C_1$ into $A_1', B_1', C_1'$. Since $\frac{IX'}{IX}=\frac{IA}{IA'}=\frac{IB}{IB'}=\frac{IY'}{IY}$ we deduce that $X'Y' \parallel XY \parallel DE$, similarity we get $Y'Z' \parallel EF$ and $Z'X' \parallel FD$ so triangle $DEF$ and triangle $X'Y'Z'$ are homothety hence $DX', EY', FZ'$ concur at one point (denote this point as $P$) ...(1)

Notice that $A_1'$ is the intersection of circle $(IX')$ and circle $(ID)$ which is obvious the projection of $I$ on $DX'$. Similarity $B_1', C_1'$ is the projection of $I$ on $EY', FZ' $. From (1) we deduce that $I, A_1', B_1', C_1', P$ are concyclic at circle $(IP)$. So $A_1, B_1, C_1$ are collinear. 

Another solution by Leo mihai Giugiuc https://www.facebook.com/leo.giugiuc?fref=ts in the pictute