AC/BC=AL/BL=AL/LB2=A2L/BL=CA1/BC=CA/CB1 => A1A2//CL//B1B2 => ∠BA1A2=∠AB1B2 BA2/sin∠BA1A2=AB2/sin∠AB1B2 => BO2=B1O1 ∠O2BA1=90-(180-∠BA2A1)=90-∠CLA=90-∠AB2B1=∠O1B1A => ⊿O2BC≌⊿O1B1C

This solution more or less is an expansion of the one above. Note that it suffices to prove that $\angle O_2CA=\angle O_1CB$. WLOG let $AC>AB$, and let $A_L=AA_1\cap CL$, $B_L=BB_1\cap CL$. It is well known that $A_1$ lies on $BC$ and $B_1$ lies on $AC$ and that $\triangle ACA_1$, $\triangle BCB_1$ are both isosceles. (The proof of this basically comes down to showing that $\triangle AA_LC=\triangle A_1A_LC$ and similarly for the triangles associated with $B_L$.) Thus $AA_1\parallel BB_1\implies AB_1=A_1B$. Next, we know that by the definition of reflection $AA_L=A_LA_1$ and $AL=LA_2$, so $CL\parallel A_1A_2$. Similarly, $CL\parallel B_1B_2$, so $A_1A_2\parallel B_1B_2$ and $\angle A_1A_2B+\angle B_1B_2A=180^\circ$. Therefore by the Extended Law of Sines, we have \[\begin{cases}\dfrac{AB_1}{\sin\angle B_1B_2A}&=2R_{(B_1B_2A)},\\ \dfrac{A_1B}{\sin\angle A_1A_2B}&=2R_{(A_1A_2B)}\end{cases}\implies R_{(B_1B_2A)}=R_{(A_1A_2B)},\] where $R_{(ABC)}$ denotes the circumradius of $\triangle ABC$. Finally, let $H_1$ and $H_2$ be the feet of the projections from $O_1$ and $O_2$ to $BC$ and $AC$ respectively. We know that $O_1A_1=O_2A$ by our work above, and in addition we know that $H_1A_1=\tfrac12(A_1B)=\tfrac12(AB_1)=H_2A$. Therefore by Pythag or HL or whatever we have $O_2H_2=O_1H_1$. Thus since $H_2A+AC=H_1A_1+A_1C\implies H_1C=H_2C$, we have that $\triangle O_2H_2C=\triangle O_1H_1C\implies \angle O_2CA=\angle O_1CB$, as desired. $\blacksquare$ What a great problem!

$AB_1=AB_2$ $B_2L=BL$ $AL=A_2L$ $B_2L-AL=BL-A_2L$ OR $B_2A=A_2B $ $m(B_2AB_1)=m(CAB)=m(A_2BA_1)$ Comparing we get $triangleB_1AB_2=triangleA_1BA_2$ $O_1A=O_2A_1$ $m(O_1AB_1)=m(O_2A_1B)$ $m(CAO_1)=m(CA_1)$ $AC=A_2C$ $triangleCAO_1=triangleCA_1O_2 [S-A-S]$ $m(O_1CA)=m(O_2CB) $ Q.E.D.

TripteshBiswas wrote: AB’=AB”…….(1) B”L=BL AL=A”L B”L-AL=BL-A”L OR B”A=A”B ……….(2) m(B”AB’)=m(CAB)=m(A”BA’)……….(3) Comparing 1,2,3 we get ∆B’AB”=∆A’BA” →O’A=O”A’ →m(O’AB’)=m(O”A’B) →m(CAO’)=m(CA’O”) AC=A’C ∆CAO’=∆CA’O” [S-A-S] →m(O’CA)=m(O”CB) Q.E.D. Your No. 3 equality is wrong (2nd part!)! Bets regards, sunken rock

$O_1$ is circumcentre of $trianleAB_1B_2$ SO $m(O_1AB_1)=90-m(B_1B_2A)$ $O_2$ is the circumcentre of $trianleA_2A_1B$ so $m(O_2A_1B)=90-m(A_1A_2B)$ $m(B_1B_2A)= m(A_1A_2B)$ gives $m(O_1AB_1)= m(O_2A_1B)$ So Please kindly tell me the detail of wrong of my solution sunken rock.

I do not understand how djmathman got this: Finally, let $H_1$ and $H_2$ be the feet of the projections from $O_1$ and $O_2$ to $BC$ and $AC$ respectively. We know that $O_1A_1=O_2A$ by our work above, and in addition we know that $H_1A_1=\tfrac12(A_1B)=\tfrac12(AB_1)=H_2A$ Can someone explain.

The equality $O_1A = O_2A$ is a result of the fact that the radii of the two circles are equal, while $H_1A_1 = \tfrac12(A_1B)$ comes from the fact that $H_1$ is the midpoint of $\overline{A_1B}$ (and similarly for $H_2A = \tfrac12(AB_1)$).

I'm surprised this solution isn't already on the thread, hopefully it isn't a fakesolve. In fact, we claim $(AB_1B_2)$ and $(BA_1A_2)$ are reflections over $CL$. It suffices to show that the reflection of $A_2$ over $CL$ lies on $(AB_1B_2)$. By homotheties at $A$ and $B$ we know $A_1A_2\parallel B_1B_2\parallel CL$. Now let $P\neq A$ be on $(AB_1B_2)$ such that $AP\parallel CL$. If we can show $AA_1A_2P$ is a rectangle, we'll be done. Since $AP\parallel B_1B_2$, $APB_1B_2$ is an isosceles trapezoid and hence the perpendicular bisector of $AP$ is also the perpendicular bisector of $B_1B_2$. But from right triangle $BB_1B_2$ this line passes through $L$, so from right triangle $AA_1A_2$ it is also the perpendicular bisector of $AA_1A_2$ and we're done.