Let $ AD $ be the bisector of a triangle $ ABC $  $ (D \in BC) $ such that $ AB + AD = CD $ and $ AC + AD = BC $. Determine the measure of the angles of $ \vartriangle ABC $
1. Given the conditions:
   \[
   \begin{cases}
   AB + AD = CD \\
   AC + AD = BC
   \end{cases}
   \]
   We need to determine the measure of the angles of $\triangle ABC$.

2. From the given conditions, we can write:
   \[
   AB + AD = CD \quad \text{and} \quad AC + AD = BC
   \]
   Subtracting the first equation from the second, we get:
   \[
   (AC + AD) - (AB + AD) = BC - CD
   \]
   Simplifying, we obtain:
   \[
   AC - AB = BC - CD
   \]
   Rearranging terms, we have:
   \[
   AB - AC = CD - BC
   \]

3. Using the angle bisector theorem, we know:
   \[
   \frac{AB}{BD} = \frac{AC}{CD}
   \]
   Let \( AB = c \), \( AC = b \), and \( BC = a \). Then:
   \[
   \frac{c}{BD} = \frac{b}{CD}
   \]

4. From the given conditions, we can express \( CD \) and \( BD \) in terms of \( a \), \( b \), and \( c \):
   \[
   CD = a + c - b \quad \text{and} \quad BD = a - c
   \]

5. Substituting these into the angle bisector theorem:
   \[
   \frac{c}{a - c} = \frac{b}{a + c - b}
   \]
   Cross-multiplying, we get:
   \[
   c(a + c - b) = b(a - c)
   \]
   Expanding and simplifying:
   \[
   ac + c^2 - bc = ab - bc
   \]
   \[
   ac + c^2 = ab
   \]
   \[
   b^2 = c^2 + ac
   \]

6. Using the trigonometric identity for the sine rule in a triangle:
   \[
   (\sin B + \sin C)(\sin B - \sin C) = \sin A \sin C = \sin (B + C) \sin C
   \]
   Simplifying further:
   \[
   4 \sin \left( \frac{B + C}{2} \right) \sin \left( \frac{B - C}{2} \right) \cos \left( \frac{B + C}{2} \right) \cos \left( \frac{B - C}{2} \right) = \sin (B + C) \sin C
   \]

7. Since \( \sin (B - C) = \sin C \), we conclude that:
   \[
   B - C = C \quad \text{or} \quad B - C = 180^\circ - C
   \]
   The first case implies \( B = 2C \), and the second case is not possible in a triangle.

8. Using the fact that the sum of angles in a triangle is \( 180^\circ \):
   \[
   A + B + C = 180^\circ
   \]
   Substituting \( B = 2C \):
   \[
   A + 2C + C = 180^\circ
   \]
   \[
   A + 3C = 180^\circ
   \]
   \[
   A = 180^\circ - 3C
   \]

9. Since \( A \) must be positive and less than \( 180^\circ \), we find:
   \[
   0 < 180^\circ - 3C < 180^\circ
   \]
   \[
   0 < 3C < 180^\circ
   \]
   \[
   0 < C < 60^\circ
   \]

10. Therefore, the angles of \( \triangle ABC \) are:
   \[
   A = 180^\circ - 3C, \quad B = 2C, \quad C = C
   \]

The final answer is \( \boxed{A = 180^\circ - 3C, B = 2C, C = C} \).