1. **Problem statement:** Given rhombus ABCD with $\angle ADC = 64^\circ$ and equilateral triangle BCE, find $\angle CAE$. 2. **Properties and formulas:** - In a rhombus, opposite angles are equal and adjacent angles are supplementary. - All sides of a rhombus are equal. - In an equilateral triangle, all angles are $60^\circ$. - We will use angle sum properties and properties of rhombus and equilateral triangle. 3. **Find $\angle DAB$:** Since ABCD is a rhombus, $\angle ADC = \angle ABC = 64^\circ$. Adjacent angles are supplementary, so $$\angle DAB = 180^\circ - 64^\circ = 116^\circ.$$ 4. **Find $\angle ABC$:** As above, $\angle ABC = 64^\circ$. 5. **Analyze triangle BCE:** Since BCE is equilateral, all angles are $60^\circ$. So, $\angle BCE = 60^\circ$. 6. **Find $\angle BCA$:** In rhombus ABCD, sides are equal, so $AB = BC = CD = DA$. Since $\angle ABC = 64^\circ$ and $\angle BCE = 60^\circ$, $\angle BCA = \angle BCD - \angle ECD$. But $\angle BCD = \angle ADC = 64^\circ$ (opposite angles in rhombus), and $\angle ECD = 60^\circ$ (from equilateral triangle). So, $$\angle BCA = 64^\circ - 60^\circ = 4^\circ.$$ 7. **Find $\angle BAC$:** In triangle ABC, sum of angles is $180^\circ$, so $$\angle BAC = 180^\circ - \angle ABC - \angle BCA = 180^\circ - 64^\circ - 4^\circ = 112^\circ.$$ 8. **Find $\angle CAE$:** Since E lies on the extension from B to E forming equilateral triangle BCE, $\angle CAE = \angle BAC - \angle BAE$. But $\angle BAE = \angle EAB = 60^\circ$ (since BCE is equilateral and shares side BC). Therefore, $$\angle CAE = 112^\circ - 60^\circ = 52^\circ.$$ **Final answer:** $$\boxed{52^\circ}.$$