<s>[INST] Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$ 

Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$

 [/INST] 
\begin{align}
x=|OA|&=|CO_1|=|A_1O_2|=|BO_3|
.
\end{align}
Let $|AD|=a$, $|DC|=2a$. 
\begin{align}
|CO|&=2a\cos\theta
,\\
|AO|=|CO_1|&=
\sqrt{(3\,a)^2+(2a\,\cos\theta)^2-2\cdot3\,a\cdot(2\,a\,\cos\theta)
 \cdot\cos\theta}
\\
&=a\,\sqrt{9-8\,\cos^2\theta}
,\\
|AO_1|&=\sqrt{|AC|^2-|CO_1|^2}
\\
&=\sqrt{(3a)^2-a^2\,(9-8\cos^2\theta)}
\\
&=2\sqrt2\,a\,\cos\theta
,\\
|OO_1|&=
\sqrt{|CO|^2-|CO_1|^2}
\\
&=\sqrt{12\,a^2\cos^2\theta-a^2\,(9-8\,\cos^2\theta)}
\\
&=a\,\sqrt{4\cos^2\theta-9}
,\\
[ACA_1B]=(3a)^2&=
2\,|CO_1|\,|AO_1|+|OO_1|^2
\\
&=
2\,a\,\sqrt{9-8\,\cos^2\theta}
\cdot 2\sqrt2\,a\,\cos\theta
+\left(a\,\sqrt{12\cos^2\theta-9}\right)^2
.
\end{align}
The last equation simplifies to
\begin{align}
-18&+4\sqrt2\sqrt{9-8\cos^2\theta}\,\cos\theta
+12\,\cos^2\theta=0
,\\
(18-12\,\cos^2\theta)^2
&=
32\,(9-8\cos^2\theta)\,\cos^2\theta
,\\
4\,(10\cos^2\theta-9)^2&=0
,\\
\cos^2\theta&=\tfrac9{10},
\\
\text{and }\quad |AO|&=
a\,\sqrt{9-8\,\cos^2\theta}
=
5\,\sqrt{9-8\cdot\tfrac9{10}}
=3\,\sqrt5
.
\end{align}
</s>