a+b & amp； #8722; c
2
, so this option is wrong;
B, let AB be tangent to F, the radius o F the circle is Y, and connect O F, as shown in Figure (2).
Then △BCA∽△OFA, ∴
Belonging to
B.C.
=
Australian second-class honorary officer
ab blood type
∴
y
a
=
b & amp#8722; y
c
The solution is: y=
abdominal muscle
a+c
, so this option is wrong;
C. connect OE, OD,
AC and BC tangent to e and d respectively,
∴∠OEC=∠ODC=∠C=90，
OE = OD，
∴ Quadrilateral OECD is a square,
∴OE=EC=CD=OD，
Let the radius of circle o be r,
∵OE∥BC,∴∠AOE=∠B，
∠∠AEO =∠0db，
∴△ODB∽△AEO，
∴
Old English
Bachelor of science
=
Automatic exposure device
mean sea level
r
a & amp#8722; r
=
b & amp#8722; r
r
Solution: r=
abdominal muscle
a+b
, so this option is correct;
Points D and O connect three tangent points, from top to bottom: OD, OE, of; Let the radius of the circle be x;
It is easy to know that BD=BF, so AD = BD-BA = BF-BA = A+X-C;
∫b-x = AE = ad = a+x-c； So x=
b+ c & amp； #8722; a
2
, this option is wrong.
So choose C.
9.。 . . . . . .
Solution: extend BC and cross the x axis at point d,
Set point C(x, y), AB=a,
∵OC bisects the angle between OA and the positive semi-axis of the X axis,
∴cd=cb′,△ocd≌△ocb′，
From the nature of folding, BC=B'C,
∵ hyperbola y = 2 x? (x > 0) passes through the vertices a, c,
∴S△OCD= 1 2 xy= 1，
∴s△ocb′= 1 2xy = 1，
BC=B'C=CD can be obtained from the property of folding transformation and the equal distance between the point on the bisector of the angle and both sides of the angle.
∴ The vertical coordinates of point A and point B are both 2y,
∫AB∨x axis,
∴ point A(x-a, 2y),
∴2y(x-a)=2，
∴xy-ay= 1，
xy = 2
∴ay= 1，
∴S△ABC= 1 2 ay= 1 2，
∴soabc=s△ocb′+s△ab'c+s△abc= 1+ 1 2+ 1 ^ 2 = 2。
So the answer is: 2.