8,9

Segment $B D$ is the median of isosceles triangle $A B C (A B = B C)$. A circle with radius 4 passes through points $B, A, D$ and intersects side $B C$ at point $E$ such that $B E: B C = 7: 8$. Find the perimeter of triangle $A B C$.
Let $CE = x$. Then $BC = 8x$ and $BE = 7x$, and since $8x = BC = AB = 8$, we have $x = 1$, $BE = 7$. Point $E$ lies on the circle with diameter $AB$, so $\angle AEB = 90^\circ$. From the right triangles $ABE$ and $ACE$, we find that

$$
AE^2 = AB^2 - BE^2 = 64 - 49 = 15, \quad AC = \sqrt{AE^2 + CE^2} = \sqrt{15 + 1} = 4.
$$

Therefore, the perimeter of triangle $ABC$ is $8 + 8 + 4 = 20$.

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