1. **State the problem:** Given positive real numbers $a$, $b$, and $c$ satisfying $$\frac{a}{a+b} = \frac{a+b}{a+b+c} = \frac{c}{b+c}$$ and the ratio $$\frac{a}{b} = \frac{1+\sqrt{x}}{y},$$ find the value of $xy$. 2. **Introduce a common ratio:** Let this common value be $k$, so $$\frac{a}{a+b} = k, \quad \frac{a+b}{a+b+c} = k, \quad \frac{c}{b+c} = k.$$ 3. **Express $a$, $b$, and $c$ in terms of $k$:** From the first equation, $$\frac{a}{a+b} = k \implies a = k(a+b) \implies a = ka + kb \implies a - ka = kb \implies a(1-k) = kb \implies \frac{a}{b} = \frac{k}{1-k}.$$ From the third equation, $$\frac{c}{b+c} = k \implies c = k(b+c) \implies c = kb + kc \implies c - kc = kb \implies c(1-k) = kb \implies c = \frac{kb}{1-k}.$$ 4. **Use the second equation relating sums:** $$\frac{a+b}{a+b+c} = k \implies a+b = k(a+b+c).$$ Rearranged, $$a+b = ka + kb + kc \\ a+b - ka - kb = kc \\ a(1-k) + b(1-k) = kc \\ (a+b)(1-k) = kc.$$ 5. **Substitute $c$ from step 3 into this:** $$ (a+b)(1-k) = k \cdot \frac{kb}{1-k} \implies (a+b)(1-k)^2 = k^2 b.$$ 6. **Express $a$ in terms of $b$ using step 3:** $$ a = \frac{k}{1-k} b, $$ so $$a + b = \frac{k}{1-k} b + b = b\left( 1 + \frac{k}{1-k} \right) = b \frac{1-k + k}{1-k} = \frac{b}{1-k}.$$ 7. **Substitute back into equation from step 5:** $$ \left( \frac{b}{1-k} \right)(1-k)^2 = k^2 b $$ Simplify left side: $$ b (1-k) = k^2 b.$$ Cancel $b$ (positive): $$1-k = k^2.$$ 8. **Rearrange and solve quadratic in $k$:** $$ k^2 + k -1 = 0.$$ Using quadratic formula, $$ k = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}.$$ Since $k$ is a ratio of positive numbers, $k$ must be positive, $$ k = \frac{-1 + \sqrt{5}}{2}.$$ 9. **Recall from step 3:** $$ \frac{a}{b} = \frac{k}{1-k}. $$ Substitute $k$, $$ \frac{a}{b} = \frac{\frac{-1 + \sqrt{5}}{2}}{1 - \frac{-1 + \sqrt{5}}{2}} = \frac{\frac{-1 + \sqrt{5}}{2}}{\frac{2 - (-1 + \sqrt{5})}{2}} = \frac{-1 + \sqrt{5}}{3 - \sqrt{5}}.$$ 10. **Rationalize denominator:** $$ \frac{-1 + \sqrt{5}}{3 - \sqrt{5}} \cdot \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{(-1 + \sqrt{5})(3 + \sqrt{5})}{(3)^2 - (\sqrt{5})^2} = \frac{(-1)(3) + (-1)(\sqrt{5}) + 3\sqrt{5} + 5}{9 - 5}.$$ Simplify numerator: $$ -3 - \sqrt{5} + 3\sqrt{5} + 5 = (-3 + 5) + ( - \sqrt{5} + 3\sqrt{5}) = 2 + 2\sqrt{5} = 2(1 + \sqrt{5}).$$ Denominator is $4$, so $$ \frac{a}{b} = \frac{2(1 + \sqrt{5})}{4} = \frac{1 + \sqrt{5}}{2}.$$ 11. **Match with given ratio form:** $$ \frac{a}{b} = \frac{1 + \sqrt{x}}{y} = \frac{1 + \sqrt{5}}{2},$$ so $$ x = 5, \quad y = 2.$$ 12. **Find $xy$:** $$ xy = 5 \times 2 = 10.$$ **Final answer:** $$\boxed{10}.$$