1. **Problem statement:** Two towers of heights 30 m and 40 m are 50 m apart. A well is located between them on the line joining their bases. Two birds fly simultaneously from the tops of the towers and land at the well at the same time, flying at the same speed. We need to find the position of the well between the towers. 2. **Express EC in terms of x:** Let the distance from point A (base of the 30 m tower) to the well E be $x$ meters. Since AC = 50 m, the distance from E to C (base of the 40 m tower) is $EC = 50 - x$. 3. **Pythagorean theorem in triangle ABE:** In triangle ABE, right-angled at A, with AB = 30 m and BE = y, $$AB^2 + AE^2 = BE^2 \implies 30^2 + x^2 = y^2 \implies 900 + x^2 = y^2.$$ 4. **Pythagorean theorem in triangle DCE:** In triangle DCE, right-angled at C, with DC = 40 m and DE = y, $$DC^2 + CE^2 = DE^2 \implies 40^2 + (50 - x)^2 = y^2.$$ Expanding, $$1600 + (50 - x)^2 = y^2,$$ $$1600 + (2500 - 100x + x^2) = y^2,$$ $$4100 + x^2 - 100x = y^2.$$ 5. **Equating the two expressions for $y^2$:** Since $DE = BE = y$, set the two expressions equal: $$4100 + x^2 - 100x = 900 + x^2.$$ Simplify by subtracting $x^2$ from both sides: $$4100 - 100x = 900,$$ $$4100 - 900 = 100x,$$ $$3200 = 100x,$$ $$x = \frac{3200}{100} = 32.$$ 6. **Conclusion:** The well is located 32 meters from the base of the 30 m tower (point A) along the line joining the two towers. **Final answer:** The well is 32 meters from the 30 m tower and therefore $50 - 32 = 18$ meters from the 40 m tower.