Let $ ABCD$ be a convex quadrilateral. We have that $ \angle BAC\equal{}3\angle CAD$, $ AB\equal{}CD$, $ \angle ACD\equal{}\angle CBD$. Find angle $ \angle ACD$
1. Let \( ABCD \) be a convex quadrilateral with \( \angle BAC = 3\angle CAD \), \( AB = CD \), and \( \angle ACD = \angle CBD \). We need to find the measure of \( \angle ACD \).

2. Assume \( AB = AD \). Let \( O \) be the intersection of diagonals \( AC \) and \( BD \). Since \( \angle DOC = \angle DBC \), we have \( \triangle DOC \sim \triangle DBC \). This implies:
   \[
   \frac{DC}{DB} = \frac{DO}{DC} \Rightarrow DC^2 = DO \cdot DB
   \]

3. Let \( x = \angle DAC \) and \( y = \angle ACD = \angle DBC \). Then, \( \angle BAC = 3x \).

4. Since \( \angle ACD = \angle CBD \), we have:
   \[
   \angle ADB = \angle ABD = 90^\circ - 2x
   \]
   and
   \[
   \angle AOD = 90^\circ + x
   \]

5. Using the Law of Sines in \( \triangle AOD \) and \( \triangle ABD \):
   \[
   \frac{\sin x}{\sin y} = \frac{\sin \angle ADC}{\sin \angle ACD} = \frac{DC}{AD}
   \]

6. Since \( AB = AD \), we have:
   \[
   \left( \frac{\sin y}{\sin x} \right)^2 = \frac{AD^2}{DC^2} = \frac{AD}{DO} \cdot \frac{AB}{DB} = \frac{\sin \angle AOD}{\sin \angle DAO} \cdot \frac{\sin \angle ADB}{\sin \angle DAB}
   \]

7. Substituting the angles:
   \[
   \left( \frac{\sin y}{\sin x} \right)^2 = \frac{\sin (90^\circ + x)}{\sin x} \cdot \frac{\sin (90^\circ - 2x)}{\sin 4x}
   \]

8. Simplifying the trigonometric expressions:
   \[
   \frac{\sin (90^\circ + x)}{\sin x} = \frac{\cos x}{\sin x} = \cot x
   \]
   and
   \[
   \frac{\sin (90^\circ - 2x)}{\sin 4x} = \frac{\cos 2x}{4 \sin x \cos x \cos 2x} = \frac{\cos 2x}{4 \sin x \cos x \cos 2x} = \frac{1}{4 \sin x}
   \]

9. Combining these results:
   \[
   \left( \frac{\sin y}{\sin x} \right)^2 = \cot x \cdot \frac{1}{4 \sin x} = \frac{1}{4 \sin^2 x}
   \]

10. Therefore:
   \[
   \sin^2 y = \frac{1}{4} \Rightarrow \sin y = \frac{1}{2}
   \]

11. Hence:
   \[
   y = 30^\circ
   \]

12. Therefore, \( \angle ACD = 30^\circ \).

The final answer is \( \boxed{30^\circ} \)