1. Let the sides of a cyclic quadrilateral $ABCD$ be $AB=a, BC=b, CD=c, DA=d$. Find the lengths of the diagonals $AC$ and $BD$ (expressed in terms of $a, b, c, d$).
1. Take a point $H$ on the arc $\overparen{A D C}$ such that $A H=C D=c$, and connect $H C, H B$. Let $A C=m, B D=n, B H=p$. It is easy to prove that $\triangle A H C \cong \triangle C D A$, which means $H C=A D=d$. Applying Ptolemy's theorem to quadrilaterals $A B C D$ and $A B C H$, we get $a c+b d=m n, a d+b c=p m$.

Then, take a point $K$ on the arc $\overparen{B C H}$ such that $B K=C H=d$. By $\triangle C H B \cong \triangle K B H$, we have $H K=B C=b$. Applying Ptolemy's theorem to quadrilateral $A B K H$, we get $a b+c d=A K \cdot p$. Since $\overparen{K H A}=\overparen{B C D}$, we have $A K=B D=n$. Therefore, $m^{2}=\frac{(a c+b d)(a d+b c)}{a b+c d}, n^{2}=\frac{(a c+b d)(a b+c d)}{a d+b c}$, from which we can find $A C, B D$.