Crux Problem 4859
Problem 4859 from Crux Mathematicorum Vol. 49, No. 6
Given a triangle \(ABC\), a point \(X\) on segment \(AB\) and a point \(Y\) on segment \(AC\), such that \(B\), \(X\), \(Y\), \(C\) are concyclic, let \(I\), \(J\), \(K\) be the incenters of triangles \(ABC\), \(XBC\), and \(YBC\), respectively. Prove that \(AI\) is orthogonal to \(JK\).
(Proposed by Trinh Quoc Khanh)
Given that \(B\), \(X\), \(Y\), \(C\) are concyclic, we have \(\angle BXC = \angle BYC \implies \angle XBC + \angle XCB = \angle YBC + \angle YCB \implies \angle BJC = \angle BKC\). Thus \(B\), \(C\), \(K\), \(J\) are concyclic,
Then we note that \(\angle XJB = 180^{\circ} - \frac{\angle BXC + \angle XBC}{2}\), \(\angle BJK = 180^{\circ} - \angle KCB\), \[ \angle XJK = 360^{\circ} - (\angle XJB + \angle BJK) = \frac{\angle BXC + \angle XBC + \angle YCB}{2} .\] Similarly \(\angle YKC = 180^{\circ} - \frac{\angle BYC + \angle YCB}{2}\), \(\angle CKJ = 180^{\circ} - \angle JBC\), \[ \angle YKJ = 360^{\circ} - (\angle YKC + \angle CKJ) = \frac{\angle BYC + \angle YCB + \angle XBC}{2} .\] Since \(\angle BXC = \angle BYC\), it follows that \(\angle XJK = \angle YKJ\).
Extend \(BJ\) and \(CK\), they will intersect at \(I\). Since \(\angle XAI = \angle YAI\), \(\angle AXJ = \angle AYK\), and given that we just showed \(\angle XJK = \angle YKJ\), we can conclude that \(AI\) is perpendicular to \(JK\). This completes the proof.