Problem
GEO-B1-M05-P034 An Angle Bisector in a Right Triangle
In right triangle \(ABC\), angle \(C\) is right. Angle bisector \(BK\) meets \(AC\) at point \(K\). The circumcircle of triangle \(ABK\) intersects line \(BC\) again at point \(L\). Prove that \(BC+CL=AB\).
C. Hint 1. On line \(BC\), beyond point \(C\), mark a segment \(CN=CL\).
D. Hint 2. It is enough to prove \(BN=AB\). Use the cyclicity of \(A,B,K,L\).
E. Full solution.
On the extension of \(BC\) beyond \(C\), mark point \(N\) so that \(CN=CL\). Then \(BN=BC+CN=BC+CL\), so it is enough to prove \(BN=AB\).
Since \(A,B,K,L\) lie on one circle, angles \(\angle AKB\) and \(\angle ALB\) are supplementary. Using the collinearities \(A,K,C\) and \(B,L,C,N\), we get \(\angle CKB=\angle ALC\).
Triangles \(ACL\) and \(ACN\) are right triangles, have common leg \(AC\), and have equal legs \(CL=CN\). They are congruent, hence \(\angle ANC=\angle ALC\). Therefore \(\angle ANB=\angle CKB\).
Since \(BK\) bisects angle \(B\), \(\angle ABK=\angle KBC\). Using the right triangle and the previous angle equality, we obtain \(\angle BAN=\angle ANB\).
Thus triangle \(ABN\) is isosceles, and \(AB=BN\). Therefore \(AB=BC+CL\).
A. Source analysis. Main objects: a right triangle, an angle bisector, and a second intersection of a circle. The obvious approach is to compute \(CL\), but the hidden construction is to lay off \(CL\) beyond the right angle and reduce the problem to an isosceles triangle. Number of key ideas: 3.
F. Difficulty justification. This is Level 6: a regional-style problem with a classical added segment and a cyclic angle.
G. Check. This is not a one-step exercise: it needs the auxiliary point \(N\), congruent right triangles, and an angle argument in \(ABN\).