Answer: 175

Small Hint:

The midpoints of all chords from $A$ lie on the circle with diameter joining $A$ to the center

Big Hint:

Uniqueness makes $BC$ tangent to that midpoint circle at the midpoint of $AD$

Solution:

Let $H$ be the midpoint of $AD.$ Put $H=(0,0),$ $A=(-a,0),$ $D=(a,0),$ and let the circle’s center be $O=(0,k).$ Then $a^2+k^2=25.$ The midpoints of all chords starting at $A$ form the circle with center $\frac{A+O}{2}$ and radius $\frac{5}{2}.$ Because $AD$ is the only such chord bisected by the line $BC,$ that line is tangent to the midpoint circle at $H.$

Hence a unit direction vector for $BC$ may be taken as $u=\frac{(k,a)}{5},$ perpendicular to $A+O=(-a,k).$ Write $B=-pu$ and $C=qu,$ where $p,q>0.$ Then $p+q=6,$ and intersecting chords give $pq=a^2.$ Substituting the line into the original circle also gives $$q-p=2u\mathbin{\cdot}O=\frac{2ak}{5}.$$ Therefore $$\begin{aligned} 36-4a^2&=(q-p)^2\\ &=\frac{4a^2(25-a^2)}{25}, \end{aligned}$$ so $a^4-50a^2+225=0.$ Its roots are $a^2=5$ and $45,$ but $a^2=pq\leq\frac{(p+q)^2}{4}=9.$ Thus $a^2=5,$ and $\{p,q\}=\{1,5\}.$

Choose $a=\sqrt5,$ $k=2\sqrt5,$ and $p=1$ for the endpoint $B$ on minor arc $AD.$ Then $$\begin{aligned} \overrightarrow{OA} &=(-\sqrt5,-2\sqrt5),\\ \overrightarrow{OB} &=\left(-\frac2{\sqrt5}, -\frac{11}{\sqrt5}\right). \end{aligned}$$ The sine of the central angle subtending minor arc $AB$ is the absolute determinant of these radius vectors divided by $5^2,$ namely $$\sin\overset{\frown}{AB}=\frac{|11-4|}{25}=\frac7{25}.$$ Hence $mn=7\cdot25=175.$