Ad split cam

∴DAC=DAE

AO = DO

∴DAC=ADO

∴ADO=DAE

∵DE⊥MN

∴DAE

ADE=90

∴ADO

ADE=90

That is ODE=90.

∴OD⊥DE

∴DE is the tangent of⊙ O.

(2) If AF⊥OD is f, then the quadrilateral AEDF is a rectangle.

∴DF=AE=3cm,AF=DE=6cm

Let the radius of ⊙O be r.

∵Rt△AOF，

∴AF

OF=AO

six

(r-3)=r

The solution is r=7.5.

DE⊥MN

Point d is the tangent point.

Connecting outer diameter

And then OD⊥ED.

So if AF⊥OD is f, then the quadrilateral AEDF is a rectangle.

Let AF⊥OD be F.

make

It can also be said that it is established