To solve this problem, let's break it down step by step using the geometric properties of circles, angles, and tangents.

Step 1: Understanding the Problem

• $O$ is the center of the circle.
• $A$ is a point on the circle, so $OA$ is the radius.
• $B$ is a point outside the circle, and $AB$ is an extended radius.
• A tangent is drawn from point $B$ to the circle, touching the circle at point $D$.
• $BD$ is perpendicular to the radius $OD$ at the point of tangency.
• $\angle AOB = 34^\circ$, and we need to find the angle $\angle OAD$.

Step 2: Analyzing the Geometry

1. Circle Center $O$, Radius $OA$:

   • $OA = OD$ because both are radii of the circle.

2. Angle Relationships:

   • Since $AB$ is a straight line, $\angle OAB = \angle AOB = 34^\circ$.
   • $BD$ is a tangent, so $\angle ODB = 90^\circ$ (because the radius $OD$ is perpendicular to the tangent at the point of tangency $D$).

Step 3: Angle $\angle OAD$

To find $\angle OAD$, note the following:

• Since $BD$ is tangent to the circle at $D$ and $OD$ is the radius, $\angle OBD$ is 90°.
• $\angle OAD$ is the angle subtended by the arc $AD$ at the point $O$.

Now, use the fact that the sum of the angles in triangle $OAB$ (considering $O$, $A$, and $B$ as vertices) is 180°:

$\angle OAB + \angle OBA + \angle AOB = 180^\circ$ Since $\angle OAB = \angle OBA$ (due to symmetry in isosceles triangle $OAB$), and $\angle AOB = 34^\circ$,

$2 \times \angle OAB + 34^\circ = 180^\circ$ $2 \times \angle OAB = 146^\circ \quad \Rightarrow \quad \angle OAB = 73^\circ$

Now, since $OA$ is the radius and $AD$ is a chord, $\angle OAD$ will be half of $\angle OAB$ because of the Inscribed Angle Theorem (angle subtended by the same chord at the center and on the circle):

[ \angle OAD = \frac{1}{2