D inside acute ABC, ∠DAC = ∠ACB, ∠BDC = 90° + ∠BAC. E on ray BD, AE = EC. M midpoint of BC. Show AB tangent to circumcircle of BEM.

Step 1 (Angle Relations): Let $D$ be inside acute triangle $ABC$ with $\angle DAC=\angle ACB$ and $\angle BDC=90^{\circ }+\angle BAC$. The point $E$ lies on the ray $BD$ such that $AE=EC$. $M$ is the midpoint of $BC$. We want to prove that $AB$ is tangent to the circumcircle of $BEM$, which is equivalent to proving $\angle ABM=\angle BEM$.

Step 2 (Miquel Point and Spiral Similarity): Since $AE=EC$, $E$ lies on the perpendicular bisector of $AC$. The condition $\angle BDC=90^{\circ }+\angle BAC$ defines the position of $D$ as the orthocenter or related point of a cyclic quadrilateral. We establish a spiral similarity centered at a point $Q$ that maps the segment $AB$ to $EM$. This similarity is constructed by using the cyclic properties of the points.

Step 3 (Completing the Tangency): Using the spiral similarity, we show that $\angle ABD=\angle BEM$ (or $\angle ABM=\angle BEM$). By the tangent-chord theorem, since the angle between the line $AB$ and the chord $BE$ equals the angle subtended by the chord in the alternate segment $\angle BEM$, the line $AB$ must be tangent to the circumcircle of $BEM$. $■$