Let angleBAE = x, and angleBCD = y.

AB = 9 cos x = AC cos 2x.

BC = 8 root 2 cos y = AC cos 2y.

Eliminating AC:

(9 cos x) / (cos 2x) = (8root 2 cos y) / (cos 2y).

y = 45° − x. Also, 2y = 90° − 2x, and so cos 2y = sin 2x. Therefore:

(9 cos x) / (cos 2x) = (8root 2 cos (45°−x)) / (sin 2x).

Using trigonometric identity cos(a − b) = cos a · cos b + sin a · sin b:

cos (45°−x) = (cos x + sin x) /root 2.

Rearranging: tan 2x = 8(cos x + sin x) / 9 cos x.

Using trigonometric identity tan 2a = 2 tan a / (1 − tan2a), and letting t = tan x:

2t / (1 − t2) = 8(1 + t)/9.

Therefore 9t/4 = (1 + t)(1 − t2) = 1 + t − t2 − t3.

Hence t3 + t2 + (5/4)t − 1 = 0.

By inspection, one root is t = 1/2.

Therefore (t − 1/2)(t2 + 3t/2 + 2) = 0.

The quadratic factor has no real roots (since (3/2)2 − 4·1·2 < 0), and so t = 1/2 is the only real root.

AC = 9 · cos x / cos 2x.

Using trigonometric identities cos x = 1 / square root(1 + t2), cos 2x = (1 − t2)/(1 + t2):

AC = 9 · square root(1 + t2) / (1 − t2).

Therefore AC = 9 · (root 5/2) / (3/4) = 6root 5 cm.