Example 8 As shown in Figure 8, in quadrilateral $ABCD$, $AB=BC=CD$, $\angle ABC=90^{\circ}$, $\angle BCD=150^{\circ}$. Find the degree measure of $\angle BAD$.
(2003, Beijing
Municipal Junior High School Mathematics Competition
(Preliminary))
As shown in Figure 8, draw a perpendicular line from point $A$ to $AB$, and then draw a perpendicular line from point $C$ to $BC$. The two perpendicular lines intersect at point $E$, and connect $DE$.

It is easy to see that quadrilateral $ABCE$ is a square, and $\triangle CDE$ is an equilateral triangle.
Thus, $AE = ED$, $\angle AEC = \angle BAE = 90^{\circ}$,
$\angle CED = 60^{\circ}$.
Therefore, $\angle AED = 150^{\circ}$, $\angle EAD = \angle EDA = 15^{\circ}$.
Hence, $\angle BAD = 90^{\circ} - \angle DAE = 75^{\circ}$.