TITLE: Finding a distance between a point in a circle from the center.
QUESTION [7 upvotes]: 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?)

REPLY [3 votes]: 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 $\triangle AOB$:
$$\begin{align}BO&=\sqrt{AO^2+AB^2-2\cdot AO\cdot AB\cdot \cos \angle OAB}=\\
&=\sqrt{50+36-2\cdot \sqrt{50}\cdot 6\cdot \frac1{\sqrt2}}=\\
&=\sqrt{26}.\end{align}$$