Geometric mean proof with two tangents There are two tangent lines on $f(x) = \sqrt{x}$ each with the $x$-value $a$ and $b$ respectively. 
I need to prove that $c$, the $x$ value of the point at which the two lines intersect each other, is equal to $\sqrt{ab}$, the geometric mean of $a$ and $b$. 
I have been trying many different ways of doing this question and I keep getting stuck. 

The tangent line at $(a,\sqrt{a})$ has equation $y=\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}$. 
The tangent line at $(b,\sqrt{b})$ has equation $y=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$.  
Set $\frac{1}{2\sqrt{a}}x+\frac{\sqrt{a}}{2}=\frac{1}{2\sqrt{b}}x+\frac{\sqrt{b}}{2}$ and solve for $x$.
