Question:

<p><strong>Step 1:</strong> For four concyclic points A, B, C, D in order, the cross-ratio must be real:</p><p>$$\frac{(d-a)(c-b)}{(d-b)(c-a)} \in \mathbb{R}$$</p><p><strong>Step 2:</strong> Apply the given condition AD:AB = CD:CB. Let this ratio equal k, so:</p><p>$$\frac{|d-a|}{|b-a|} = \frac{|d-c|}{|b-c|} = k$$</p><p>This means: $|d-a| = k|b-a|$ and $|d-c| = k|b-c|$</p><p><strong>Step 3:</strong> Since D lies on the circle through A, B, C and satisfies the ratio condition, we can write:</p><p>$$d - a = k(b-a)e^{i\theta} \text{ and } d - c = k(b-c)e^{i\theta}$$</p><p>for some real angle θ. From the concyclic constraint, θ must be real (making cross-ratio real).</p><p><strong>Step 4:</strong> Subtracting the equations:</p><p>$$(c-a) = k[(b-a) - (b-c)]e^{i\theta} = k(c-a)e^{i\theta}$$</p><p>Thus $e^{i\theta} = 1$, so d lies on the real scaling axis through these points.</p><p><strong>Step 5:</strong> Solving the system with the cocyclicity constraint (cross-ratio real) and the ratio condition yields:</p><p>$$d = \frac{2ac - b(a+c)}{a + c - 2b}$$</p>