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On sides $AC$ and $BC$ of $\triangle ABC$ squares $ACA_1A_2$ and $BCB_1B_2$ are constructed outwards. Prove that lines $A_1B, A_2B_2$, and $AB_1$ are concurrent. enter image description here

The only thing I have tried is using the fact that $ACA_1A_2$ and $BCB_1B_2$ are cyclic. I am not sure where to go from here.

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By a rotation around $C$ we get that $BA_1$ is congruent an orthogonal to $B_1A$.
Let $X=BA_1\cap AB_1$: such point lies on the circle with diameter $AA_1$ and on the circle with diameter $BB_1$. The bisector of $\widehat{AXA_1}$ goes through $A_2$ by equal arcs/chords in a circle, and the bisector of $\widehat{BXB_1}$ (that is the same line) goes through $B_2$ by equal arcs/chords in another circle.
Done.

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