Question:
Given a $\triangle ABC$, a pair of congruent circles are drawn(one pair at a time, in the figure colored Red,Orange,Blue) by taking $B,C$ as centers. They cut sides $BC,AC$ at pair of points $($in fig. $(D,G);(E,H);(F,I)$$)$.
Now taking those pair of points and $A$ as vertices of triangles, circumcircles are drawn(in the fig. green colored.) Prove that they all pass through a common point which is $(X$ in the figure$)$
Figure:
The question was long, I solved it till here(where I'm stuck.)
Thanks for sparing time to view my question and providing hints,solutions.
Let $U,V$ be points on $BA,CA$ such that $BU=CV$. Let $M$ be the midpoint of $BC$ and let $X$ be the midpoint of the major $BC$-arc in the circumcircle of $ABC$.
My claim is that $A,U,V,X$ belong to the same circle, i.e. $\widehat{UXV}=\widehat{BAC}=\widehat{A}$.
It is simple to notice that $\widehat{UBX}=\widehat{VCX}$, hence $UBX=VCX$ by SAS.
Indeed $\widehat{XBC}=\widehat{XCB}=\frac{\pi-\widehat{A}}{2}$ and $$ \widehat{UBX}=\widehat{B}-\frac{\pi-\widehat{A}}{2} = \frac{\pi-\widehat{A}}{2}-\widehat{C}=\widehat{VCX}.$$ This gives $$ \widehat{UXV}=\widehat{BXC}=\widehat{A}$$ as wanted.