ABC is a right triangle,
So the radius of the inscribed circle r=(6+8- 10)/2=2.
So S⊙o=4π.
(2) connect OE, OF,
Because AB and AC are tangents of circle O.
So OE⊥AC OF⊥AB
So a, e, d and f are quadrilaterals inscribed in a circle.
So ∠ A+∠ EO ∠ F = 180.
∠EDF= 1/2∠EOF。
So EDF = 90- 1/2 * 88 = 46.
(3) According to (2)
∠EDF=90 - 1/2*∠A
∠FED=90 - 1/2*∠B
∠DFE=90 - 1/2*∠C
So the triangle DEF must be an acute triangle.