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Problem 2. Given a triangle $ABC$, let $K$ and $L$ be distinct points on side $AC$ such that $\angle ABK = \angle CBL$. Rays $BK$ and $BL$ are not orthogonal to $AC$, and intersect the circumcircle of triangle $ABC$ for the second time at points $K_1$ and $L_1$, respectively. Points $K_2$ and $L_2$ lie on the tangents to the circumcircle of triangle $ABC$ at points $K_1$ and $L_1$, respectively, such that $\angle BKK_2 = \angle BLL_2 = 90^{\circ}$.
Prove that the points $A, C, K_2,$ and $L_2$ lie on a circle.
Solution 1Solution 2Solution 3Solution 4
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