Ptolemy's Theorem

Step 1: Construct an auxiliary point. On diagonal $AC$, construct point $K$ such that $\angle ABK = \angle DBC$ (i.e. ray $BK$ makes with $BA$ the same angle that $BD$ makes with $BC$), with $K$ chosen on segment $AC$.

Step 2: Find a first pair of similar triangles. Since $ABCD$ is cyclic, $\angle BAC = \angle BDC$ (both subtend arc $BC$). Combined with $\angle ABK = \angle DBC$ by construction, triangles $ABK$ and $DBC$ share two equal angles, so $\triangle ABK \sim \triangle DBC$ by AA.

Step 3: Extract the first relation. From $\triangle ABK \sim \triangle DBC$: $\frac{AK}{DC} = \frac{AB}{DB}$, so $AK \cdot DB = AB \cdot DC$.

Step 4: Find a second pair of similar triangles. Since $ABCD$ is cyclic, $\angle ABD = \angle ACD$ (both subtend arc $AD$). Consider triangles $KBC$ and $ABD$: $\angle KBC = \angle ABC - \angle ABK = \angle ABC - \angle DBC = \angle ABD$ (since $K$ lies on segment $AC$), and $\angle BCK = \angle BCA = \angle BDA$ (both subtend arc $AB$), so $\triangle KBC \sim \triangle ABD$ by AA.

Step 5: Extract the second relation. From $\triangle KBC \sim \triangle ABD$: $\frac{KC}{AD} = \frac{BC}{BD}$, so $KC \cdot BD = AD \cdot BC$.

Step 6: Add the two relations. Adding the results of Steps 3 and 5: $AK\cdot DB + KC\cdot BD = AB\cdot DC + AD\cdot BC$, i.e. $BD(AK+KC) = AB\cdot DC + AD\cdot BC$. Since $K$ lies on segment $AC$, $AK+KC=AC$, giving exactly $AC\cdot BD = AB\cdot CD + AD\cdot BC$.