Because DE⊥BE,

And DE is the tangent of the circle, so OD⊥DE.

Okay, then.

In Δ δACB, O is the midpoint of AB, OD//BE.

Therefore, point F is the midpoint between AC and OD⊥AC(AC⊥BC,OD//BE).

So AD=DC

Because AD +DC=40

So AD=DC=20.

In δAOD, OA=OD=50/3. AD=20。 Its area is determined by Helen's formula S=√[p(p-a)(p-b)(p-c)] where

p =( 1/2)(a+b+c)=( 1/2)(20+50/3+50/3)= 80/3

Then s = √ [(80/3) (80/3-50/3) (80/3-50/3) (80/3-20)]

=400/3

s =( 1/2)odx af =( 1/2)(50/3)xaf。

Then 25/3AF=400/3.

AF= 16

So AC=2AF=32.

Then BC = √ (AB 2-AC 2)

=√[( 100/3)^2-32^2]

=28/3