Problem
GEO-B2-M01-P122 A Parallel to the Tangent
Triangle \(ABC\) is inscribed in a circle \(\Omega\). The tangent to \(\Omega\) at \(A\) is parallel to the line through \(B\) meeting \(AC\) at \(D\). Prove that \(AB^2=AC\cdot AD\).
C. Hint 1. Replace the angle between the tangent and \(AB\) by angle \(ACB\).
D. Hint 2. After using parallelism, find two equal angles in triangles \(ABD\) and \(ACB\).
E. Full solution. Let \(l\) be the tangent to \(\Omega\) at \(A\). By the statement, \(BD\parallel l\).
By the tangent-chord theorem, the angle between \(l\) and \(AB\) equals \(\angle ACB\). Since \(BD\parallel l\), we get \(\angle ABD=\angle ACB\).
Also \(D\in AC\), so \(\angle BAD=\angle BAC\). Therefore \(\triangle ABD\sim\triangle ACB\) by two angles.
From the similarity, \(\frac{AB}{AC}=\frac{AD}{AB}\). Multiplying gives \(AB^2=AC\cdot AD\), as required.
Method comment. the line parallel to the tangent transfers the tangent-chord theorem inside the 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 3 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.