Geometry · Angle bisector · Altitude · Isosceles triangle · Cyclic quadrilateral · Orthocenter · Perpendicular bisector
Let \(ABC\) be an acute triangle. The bisector of its interior angle at \(B\) meets \(AC\) at \(K\), and \(CD\) is the altitude from \(C\), with \(D\) on \(AB\). Let \(N\) be the point of \(CD\) for which \(KN \perp BC\), and let \(M\) be the intersection point of \(BK\) and \(CD\). The circle through \(B\), \(K\) and \(N\) meets the line \(AB\) a second time at \(P\), so \(P \neq B\). Prove that \(PK = PM\).
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Serbian Republic Competition (Republicko takmicenje) 2001, high school grade I, category A, problem 2. Organized by the Mathematical Society of Serbia (DMS). Source