4. (3 points) On the side $AC$ of triangle $ABC$, a circle with a radius of 10 cm is constructed with $AC$ as its diameter. This circle intersects sides $AB$ and $BC$ at points $X$ and $Y$ respectively. Find $AX \cdot AB + CY \cdot BC$.
Answer: 400

Solution: Let point $M$ be the midpoint of side $AC$, which is also the center of the circle. Then $AB \cdot BX = BC \cdot BY = (BM + 10)(BM - 10)$. Therefore, $BX = \frac{BM^2 - 100}{AB} = \frac{2AB^2 + 2BC^2 - 400 - 400}{4AB}$. Accordingly, $AX = AB - \frac{2AB^2 + 2BC^2 - 800}{4AB} = \frac{2AB^2 - 2BC^2 + 800}{4AB}$.

Similarly, $CY = \frac{2BC^2 - 2AB^2 + 800}{4BC}$. Then $AX \cdot AB + CY \cdot BC = \frac{1600}{4} = 400$.