, 2010

Let $r$ be the radius of the circle, and let $a'$ and $b'$ be the respective lengths of $PA$ and $PB$. Since $b' = 2r \sin \angle PAB = 2rc/a'$, $c = a'b'/(2r)$. Let $AC$ be the diameter of the circle and $H$ the foot of the perpendicular from $P$ to $AC$. The similarity of the triangles $ACP$ and $APH$ imply that $AH : AP = AP : AC$ or $(a')^2 = 2ra$. Similarly, $(b')^2 = 2rb$. Hence
$$c^2 = \frac{(a')^2}{2r} - \frac{(b')^2}{2r} = ab$$
as desired.

Let $E$, $F$, $G$ be the feet of the perpendiculars to the tangents at $A$ and $B$ and the chord $AB$, respectively. We need to show that $PE : PG = PG : GF$, where $G$ is the foot of the perpendicular from $P$ to $AB$. This suggests that we try to prove that the triangles $EPG$ and $GPF$ are similar.

Since $PG$ is parallel to the bisector of the angle between the two tangents, $\angle EPG = \angle FPG$. Since $AEPG$ and $BFPG$ are concyclic quadrilaterals (having opposite angles right), $\angle PGE = \angle PAE$ and $\angle PFG = \angle PBG$. But $\angle PAE = \angle PBA = \angle PBG$, whence $\angle PGE = \angle PFG$. Therefore triangles $EPG$ and $GPF$ are similar.

The argument above with concyclic quadrilaterals only works when $P$ lies on the shorter arc between $A$ and $B$. The other case can be proved similarly.