3. As shown in Figure 16, quadrilateral $ABCD$ is a square with side length $a$. A circular arc centered at $D$ with radius $DA$ intersects a semicircle with diameter $BC$ at another point $P$. Extending $AP$ intersects $BC$ at point $N$. Then $\frac{BN}{NC}=$ $\qquad$
(2004, National Junior High School Mathematics League)
Hint: The problem involves the basic figures of overlapping and hidden intersecting circles, common tangents, similarity, and projections. Extend $C P$ and $D A$ to meet at point $F$, and connect $B P$. Answer: $\frac{1}{2}$.

