Geometry · Incenter · Angle bisectors · Cyclic quadrilateral · Inscribed angle
In triangle \(ABC\) the angle at \(B\) equals \(60^\circ\). The bisector of \(\angle CAB\) meets the opposite side at \(D\), the bisector of \(\angle BCA\) meets the opposite side at \(E\), and \(S\) is the centre of the inscribed circle of the triangle. Prove that \(SD = SE\).
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Serbian Municipal Competition 2002, high school grade I, category A, problem 2. Organized by the Mathematical Society of Serbia (DMS). Source