$\overline{AB}$ is a chord of a circle with centre $O$. Line $l$ touches the circle at $B$. The foot of the perpendicular from $A$ to $l$ is $D$. Prove that $\angle BAO \cong \angle BAD$.

(N/A) Given: $\overline{AB}$ is a chord of a circle with centre $O$. Line $l$ is a tangent to the circle at $B$. $D$ is the foot of the perpendicular from $A$ to line $l$ (i.e.,$\overline{AD} \perp l$).
To prove: $\angle BAO \cong \angle BAD$.
Proof:
$1$. In $\Delta OAB$,$\overline{OA} \cong \overline{OB}$ (radii of the same circle).
$2$. Therefore,$\angle ABO \cong \angle BAO$ (angles opposite to equal sides are equal) ... $(1)$.
$3$. Line $l$ is a tangent to the circle at $B$,and $\overline{OB}$ is the radius through the point of contact $B$. Therefore,$\overline{OB} \perp l$.
$4$. We are given that $\overline{AD} \perp l$.
$5$. Since both $\overline{OB}$ and $\overline{AD}$ are perpendicular to the same line $l$,they must be parallel to each other $(\overline{OB} \parallel \overline{AD})$.
$6$. Considering $\overline{OB} \parallel \overline{AD}$ with $\overleftrightarrow{AB}$ as a transversal,the alternate interior angles are equal.
$7$. Therefore,$\angle ABO \cong \angle BAD$ (alternate interior angles) ... $(2)$.
$8$. From equations $(1)$ and $(2)$,since both $\angle BAO$ and $\angle BAD$ are equal to $\angle ABO$,we conclude that $\angle BAO \cong \angle BAD$.