Finding a distance between a point in a circle from the center. Given a diagram like this,

Where $O$ is the center and $OA = \sqrt{50}$, $AB = 6$, and $BC = 2$. The question was to find the length of $OB$. $\angle ABC = 90^o$
What I've done is so far:
I made the triangle $ABC$ and named $\angle BAC = \alpha$ . By trigonometry, I have the values for $\sin{\alpha}$ and $\cos{\alpha}$. I get $\cos{\alpha}=\frac{6}{\sqrt{40}}$.
Then I made the triangle $OCA$ and named $\angle OAB = \beta$ so $\angle OAC = \alpha + \beta$. By using the cosinus rule, I have $\cos(\alpha + \beta) = \frac{1}{\sqrt{5}}$.
Using the formula, $\cos(\alpha + \beta) = \cos{\alpha}.\cos{\beta} - \sin{\alpha}.\sin{\beta}$ and making $\sin{\beta} = \sqrt{1 -\cos^2{\beta}}$ I finally get that $\cos{\angle OAB} = \frac{1}{\sqrt{2}}$.
Finally, by using the cosinus rule on the triangle $AOB$ I get $OB = \sqrt{26}$.
My only problem is this takes me way too long!  I am interested in a quicker way to do this (i.e. I now know that $\angle OAB = 45^o$ from trigonometry, but is there a quicker way to recognize it?)
 A: Assuming $\angle ABC=90^o$ is given.
You can get there slightly quicker:
By Pythagoras, $|AC|=\sqrt{40}$.
$OAC$ is isosceles, with $|OA|=|OC|=\sqrt{50}$.
You can then immediately get $\cos(\angle OAC)=\frac{|AC|/2}{|OA|} = \frac{\sqrt{40}/2}{\sqrt{50}}= \frac{1}{\sqrt{5}}$.
I don't yet see a way to shortcut the rest.
You could do it completely differently, by algebra. Use a coordinate system, centred on $B$, and let $O$ be the point $(x,y)$. Then we get two equations from the fact that $|OA|=|OC|=\sqrt{50}$.
$$x^2+(6-y)^2=50\\
(2-x)^2+y^2=50$$
There are fairly easily solved to give $y=1$, $x=-5$, from which you get $|OB|=\sqrt{26}$.
A: Refer to the figure:
$\hspace{4cm}$
From the right triangle $ACD$: $CD=\sqrt{AD^2-AC^2}=4\sqrt{10}$.
From similarity of right triangles $ABC$ and $CDE$:
$$\frac{CE}{AB}=\frac{CD}{AC}\Rightarrow CE=12\\
DE=\sqrt{CD^2-CE^2}=4=BF\\
BE=CE-BC=12-2=10=DF=AF$$
Hence, $\angle DAF=45^\circ=\angle OAB$, indeed.
Finally, from the cosine theorem for $\triangle AOB$:
$$\begin{align}BO&=\sqrt{AO^2+AB^2-2\cdot AO\cdot AB\cdot \cos \angle OAB}=\\
&=\sqrt{50+36-2\cdot \sqrt{50}\cdot 6\cdot \frac1{\sqrt2}}=\\
&=\sqrt{26}.\end{align}$$
A: 
A slight variation of the solution
Note that $R$ is circumradius of $\triangle ADC$,
\begin{align}
|CD|&=2R\sin\alpha=2\sqrt5
,\\
|BD|&=\sqrt{|CD|^2-a^2}=4
,\\
|AD|&=c+BD=10
.
\end{align}
By the Stewart’s Theorem
for $\triangle AOD$,
\begin{align} 
|OD|^2\cdot c+|OA|^2\cdot |BD|
-|AD|\cdot(|OB|^2+c\cdot |BD|)
&=0
,\\
|AD|\cdot (
R^2
-|OB|^2-c\cdot |BD|)
&=0
,
\end{align}
\begin{align} 
|OB|^2&=
R^2-c\cdot |BD|
\\
&=50-6\cdot4
=26
.
\end{align}
