9. In cyclic quadrilateral $A B C D$ with $A B=A D=49$ and $A C=73$, let $I$ and $J$ denote the incenters of triangles $A B D$ and $C B D$. If diagonal $\overline{B D}$ bisects $\overline{I J}$, find the length of $I J$.
Answer: $\frac{28}{5} \sqrt{69}$
Let $O$ be circumcenter, $R$ the circumradius and $r$ the common inradius. We have $I O^{2}=J O^{2}=$ $R(R-2 r)$ by a result of Euler; denote $x$ for the common value of $I O$ and $J O$. Additionally, we know $A J=A B=A D=49$ (angle chase to find that $\angle B J A=\angle J B A)$. Since $A$ is the midpoint of the arc $\widehat{B D}$ not containing $C$, both $J$ and $A$ lie on the angle bisector of angle $\angle B C D$, so $C, J, A$ are collinear. So by Power of a Point we have
$$
R^{2}-x^{2}=2 R r=A J \cdot J C=49 \cdot 24 .
$$

Next, observe that the angle bisector of angle $B A D$ contains both $I$ and $O$, so $A, I, O$ are collinear. Let $M$ be the midpoint of $I J$, lying on $\overline{B D}$. Let $K$ be the intersection of $I O$ and $B D$. Observing that the right triangles $\triangle I M O$ and $\triangle I K M$ are similar, we find $I M^{2}=I K \cdot I O=r x$, so $I J^{2}=4 r x$. Now apply Stewart's Theorem to $\triangle A O J$ to derive
$$
R(x(R-x)+4 r x)=49^{2} x+x^{2}(R-x) .
$$

Eliminating the common factor of $x$ and rearranging gives
$$
49^{2}-(R-x)^{2}=4 R r=48 \cdot 49
$$
so $R-x=7$. Hence $R+x=\frac{49 \cdot 24}{7}=168$, and thus $2 R=175,2 x=161$. Thus $r=\frac{49 \cdot 24}{175}=\frac{168}{25}$.
Finally, $I J=2 \sqrt{r x}=2 \sqrt{\frac{84 \cdot 161}{25}}=\frac{28 \sqrt{69}}{5}$.