enter image description hereenter image description here Circle with radius r touches both altitudes, diameter and arc of semicircle.
Prove $r=\sqrt{ab}$ or square root of product of a and b. I have used $\sin(A+B)$, formula cosine rule but couldn't solve it. Please help and cope up with the drawing I think it's understandable.
With the notations of the figure (which takes into account the fundamental fact that two tangent circles are such that their centers are aligned with their point of tangency):
Pythagoras theorem applied to the blue right triangle gives:
$$(R-s)^2=c^2+s^2 \ \iff \ c^2=(R-s)^2-s^2\tag{1}$$
We read on the figure:
$$a=R+c-s, \ \ b=R-(c+s)$$
which implies :
$ab=(R-s)^2-c^2 \tag{2}$
Using (1) in (2), we get: $ab=s^2$
which indeed means that $$s=\sqrt{ab}$$.
Edit: I have included (green curve) the locus of the centers of the circles. It is a parabola with equation:
$$y=\frac12(1-x^2)$$
(in the given system of coordinates)
