Consider a rectangle ABCD with BC = 2AB. Let ω be the circle that touches the sides AB, BC, and AD. A tangent drawn from point C to the circle ω intersects the segment AD at point K. Determine the ratio AK/KD.
Without loss of generality the the rectangle be 2 by 4, so then the circle has unit radius.
Let O be the center of the circle
Let L be where CK is tangent to the circle and M be where BC is tangent to the circle. Then CM=3.
OMC and OLC are congruent right triangles.
Let q be angle MCO, then LCO is also q and angle KCD is 90-2q.
tan(q) = 1/3 from either right triangle OMC or OLC.
tan(90-2q) = KD/2, then KD=2*tan(90-2q)
tan(90-2q) = 1/tan(2q) = [1-(tan (q))^2]/[2*tan(q)] = [1-(1/3)^2]/[2*(1/3)] = [8/9]/[2/3] = 12/9 = 4/3.
Then KD=2*4/3=8/3, and then AK=4=8/3 = 4/3
Finally AK/KD = (4/3)/(8/3) = 1/2.
Solution
