11.24. Given an angle $X A Y$ and a point $O$ inside it. Draw a line through point $O$ that cuts off a triangle of the smallest area from the given angle.
11.24. Consider the angle \(X^{\prime} A^{\prime} Y^{\prime}\), symmetric to the angle \(X A Y\) with respect to point \(O\). Let \(B\) and \(C\) be the points of intersection of the sides of these angles. Denote the points of intersection of the line passing through point \(O\) with the sides of the angles \(X A Y\) and \(X^{\prime} A^{\prime} Y^{\prime}\) as \(B_{1}, C_{1}\) and \(B_{1}^{\prime}, C_{1}^{\prime}\) respectively (Fig. 11.5). Since \(S_{A B_{1} C_{1}} = S_{A^{\prime} B_{1}^{\prime} C_{1}^{\prime}}\), then \(S_{A B_{1} C_{1}} = (S_{A B A^{\prime} C} + S_{B B_{1} C_{1}^{\prime}} + S_{C C_{1} B_{1}^{\prime}}) / 2\). The area of triangle \(A B_{1} C_{1}\) is minimal if \(B_{1} = B\) and \(C_{1} = C\), i.e., the desired line is \(B C\).