Circle $O$ is tangential to triangle $ABC$ at $B$ and passes $C$. Express $AO$ as $\frac{a√b}{c√d}$ Let $ABC$ be a triangle with side lengths $AB = 7, BC = 8, AC = 9$. Draw a circle tangent to $AB$ at $B$ and passing through $C$. Let the center of the circle be $O$. The length of $AO$ can be expressed as $\frac{a√b}{c√d}$ for positive integers a, b, c, d where gcd(a, c) = gcd(b, d) = 1 and b, d are not divisible by the square of any prime. Find a + b + c + d.
I drew everything and cant find any patterns


According to the cosine rule
$$\cos B = \frac{7^2+8^2-9^2}{2\cdot 7\cdot8}=\frac27,\>\>\>\>\>\sin B= \frac{3\sqrt5}7
$$
Then, the radius $OB$ is
$$OB = \frac{BC}{2\cos\angle OBC}=\frac4{\sin B}=\frac{28}{3\sqrt5}
$$
Per the Pythagoras' Theorem
$$AO=\sqrt{AB^2+BO^2}=\sqrt{7^2+ \frac{28^2}{45}}=\frac{7\sqrt{61}}{3\sqrt5}
$$
Thus, $a+b+c+d=76$.
