In the triangle ABC, the angle c = 90 degrees, the circle with a point o on AB as the center and the length of OA as the radius is tangent to BC at points D, AC and ab at points E and f respectively (1) If AC = 6, ab = 10, find the radius of ⊙ o; (2) connect OE, ED, DF, EF, if the quadrilateral bdef is a parallelogram, try to judge the shape of ofDe and explain the reason

In the triangle ABC, the angle c = 90 degrees, the circle with a point o on AB as the center and the length of OA as the radius is tangent to BC at points D, AC and ab at points E and f respectively (1) If AC = 6, ab = 10, find the radius of ⊙ o; (2) connect OE, ED, DF, EF, if the quadrilateral bdef is a parallelogram, try to judge the shape of ofDe and explain the reason

(1) Because angle c = 90 degrees, OD ⊥ BC
So od / / AC,
OD/AC=OB/AB
Let ⊙ o radius = R, that is, OD = OA = of = OE = R
And AC = 6, ab = 10, so BC = 10
So R / 6 = (10-r) / 10
The solution is: r = 15 / 4
(2) If the quadrilateral bdef is a parallelogram, EF = BD = 2CD, that is Bo = 2ao
So fo = FB = ed and of / / ed, so ofDe is a parallelogram
Since of = OE, ofDe is a diamond