4. (3 points) On the side $B C$ of triangle $A B C$ as a diameter, a circle with a radius of 20 cm is constructed. This circle intersects sides $A B$ and $A C$ at points $X$ and $Y$ respectively. Find $B X \cdot A B + C Y \cdot A C$.
Answer: 1600

Solution: Let point $M$ be the midpoint of side $B C$, which is also the center of the circle. Then $A B \cdot A X = A C \cdot A Y = (A M + 20)(A M - 20)$. Therefore, $A X = \frac{A M^{2} - 400}{A B} = \frac{2 A B^{2} + 2 A C^{2} - 1600 - 1600}{4 A B}$. Accordingly, $B X = A B - \frac{2 A B^{2} + 2 A C^{2} - 3200}{4 A B} = \frac{2 A B^{2} - 2 A C^{2} + 3200}{4 A B}$.

Similarly, $C Y = \frac{2 A C^{2} - 2 A B^{2} + 3200}{4 A C}$. Then $B X \cdot A B + C Y \cdot A C = \frac{6400}{4} = 1600$.