+207 In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $1$ and $BC = \sqrt{3}$, then find $AB$.
+129847 
Construct a circle at B with radius BC that intercepts the semi-circle at E
Because BC = BE......then BE will be tangent to the semi-circle
Carry BE through to A ....let AD = x
OE = OC
BC = BE
BO = BO
Then triangles EOB and COB are congruent right triangles
tan angle OBC = 1/sqrt 3 = tan angle OBE
tan ( OBC + OBE) = tan (ABC)....so.....
[ 2 * 1/sqrt 3 ] / [1 - (1/sqrt 3)^2] = (2 + x) / sqrt 3
2/sqrt 3 * sqrt 3 = ( 1 -1/3) (2 + x)
2 = (2/3) (x + 2)
3 = x + 2
x =1
So
AB = sqrt [ AC^2 + BC^2 ] = sqrt [ (2 + x)^2 + (sqrt3)^2 ] = sqrt [ 3^2 + 3 ] = sqrt 12 = 2sqrt 3
