Problem
GEO-B2-M01-P114 Tangents and a Hidden Orthocenter
An acute triangle \(ABC\) is inscribed in a circle \(\Omega\). The tangents to \(\Omega\) at \(B\) and \(C\) meet at \(P\). From \(P\), perpendiculars \(PD\) and \(PE\) are dropped to lines \(AB\) and \(AC\), respectively. Point \(M\) is the midpoint of \(BC\). Prove that \(M\) is the orthocenter of triangle \(ADE\).
C. Hint 1. Use \(PB=PC\), hence \(PM\perp BC\).
D. Hint 2. Prove that \(M,C,E,P\) and \(M,D,B,P\) are cyclic, then apply the tangent-chord theorem.
E. Full solution. Since the tangent segments are equal, \(PB=PC\). Hence triangle \(PBC\) is isosceles, so the median \(PM\) to base \(BC\) is an altitude: \(PM\perp BC\).
Because \(PE\perp AC\), we have \(\angle PEC=90^\circ\). Also \(\angle PMC=90^\circ\). Therefore \(M,C,E,P\) lie on one circle.
In this circle, \(\angle MEP=\angle MCP\), since both angles subtend chord \(MP\). But \(CP\) is tangent to \(\Omega\) at \(C\), so by the tangent-chord theorem \(\angle MCP=\angle BAC\).
Since \(PE\perp AC\), we get \(\angle MEA=90^\circ-\angle MEP=90^\circ-\angle BAC\). Also \(\angle DAE=\angle BAC\), because \(D\in AB\) and \(E\in AC\). Thus \(\angle MEA+\angle DAE=90^\circ\), so \(ME\perp AD\).
Similarly, using the cyclic quadrilateral \(M,D,B,P\) and the tangent \(PB\), we obtain \(MD\perp AE\). Thus in triangle \(ADE\), the lines \(ME\) and \(MD\) are altitudes. Their intersection is \(M\), so \(M\) is the orthocenter of triangle \(ADE\).
Method comment. quadrilaterals \(MCEP\) and \(MDBP\) are cyclic, and the tangent converts angles at \(C\) and \(B\) into angles of the original triangle Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.
If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.
F. Difficulty justification. This is Level 7: a strong regional-style problem. It contains not only equal angles, but also a choice of an auxiliary configuration: a circle, a Miquel point, an orthic configuration, or a tangent. Without that first choice, the solution does not start.
G. Check. This is not a one-step exercise: it requires 4 key ideas. First one must recognise the hidden geometric structure, then make an angle replacement or add a circle, and only after that complete the final conclusion.