Calculate the CAD Angle in an Isosceles Triangle with Height AD

To solve this problem, we'll apply the properties of isosceles and right triangles.

• Step 1: Recognize that since $\triangle ABC$ is isosceles, angles $\angle \text{BAD} = \angle \text{CAD}$, and we are given $\angle \text{ABD} = 55^\circ$.
• Step 2: In $\triangle ABD$, use the angle sum property for a triangle, $\angle \text{BAD} + \angle \text{ABD} + \angle \text{BDA} = 180^\circ$.
• Step 3: Given that $\angle \text{BDA} = 90^\circ$ (since $AD$ is the height), calculate $\angle \text{BAD}$.

Applying the angle sum property:
$\angle \text{BAD} + 55^\circ + 90^\circ = 180^\circ$

Simplifying, we find:
$\angle \text{BAD} = 180^\circ - 145^\circ$
$\angle \text{BAD} = 35^\circ$

Thus, the angle $\angle \text{CAD}$ is also $35^\circ$ because $\triangle ABC$ is isosceles and $\angle \text{BAD} = \angle \text{CAD}$.

Therefore, the solution to this problem is $35$.