Problem
GEO-B2-M01-P120 A Miquel Point Without a Hint
In triangle \(ABC\), angle \(A\) is \(69^\circ\). Point \(D\) lies on side \(AB\), point \(E\) lies on side \(AC\), with \(AD:DB=3:4\) and \(AE:EC=6:8\). Lines \(BE\) and \(CD\) meet at \(P\). Circles \((BDP)\) and \((CEP)\) meet again at \(M\). Prove that \(A,B,E,M\) lie on one circle, and find \(\angle BME\).
C. Hint 1. The ratios define the positions of the points, but the main move is angular.
D. Hint 2. Split \(\angle BME\) as \(\angle BMP+\angle PME\).
E. Full solution. Since \(B,D,P,M\) lie on one circle, \(\angle BMP=\angle BDP\): both angles subtend chord \(BP\).
Points \(B,D,A\) are collinear, and points \(D,P,C\) are collinear. Hence \(\angle BDP\) is the oriented angle between lines \(AB\) and \(CD\).
Similarly, from the cyclic quadrilateral \(C,E,P,M\), we get \(\angle PME=\angle PCE\). But \(P,C,D\) are collinear and \(C,E,A\) are collinear, so \(\angle PCE\) is the oriented angle between \(CD\) and \(AC\).
Adding these two angles, we see that \(\angle BME\) equals the angle between \(AB\) and \(AC\), that is \(\angle BAC=69^\circ\).
But \(\angle BAE=\angle BAC\), since \(E\in AC\). Therefore \(\angle BME=\angle BAE\). By the converse criterion for equal inscribed angles, \(A,B,E,M\) lie on one circle.
Method comment. the second intersection of the two circles is a Miquel point and must also lie on the circle through \(A,B,E\) 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.