Can you find the length of this line?

Right triangles $BDO$ and $BEO$ are congruent, so $\angle DBO = \angle EBO$ . Hence, $BO$ bisects $\angle DBE$ .

Then by angle bisector theorem, $\frac{AO}{CO} = \frac{AB}{BC} = \frac{12}{18} = \frac{2}{3},$ so $AO = \frac{2}{5} \cdot AC = 10.$

//Apply Cosine Rule for $\angle A$ and $\angle C$ Cosine Rule

$\cos { \left( \angle A \right) } =\frac { { 12 }^{ 2 }+{ 25 }^{ 2 }-{ 18 }^{ 2 } }{ 2\times 12\times 25 } =\frac { 89 }{ 120 }$

$\cos { \left( \angle C \right) } =\frac { { 18 }^{ 2 }+{ 25 }^{ 2 }-{ 12 }^{ 2 } }{ 2\times 18\times 25 } =\frac { 161 }{ 180 }$

//Apply $\sin { \left( \theta \right) } =\sqrt { 1-\cos ^{ 2 }{ \left( \theta \right) } }$

$\sin { \left( \angle A \right) } =\sqrt { 1-\cos ^{ 2 }{ \left( \angle A \right) } } =\frac { \sqrt { 6479 } }{ 120 }$

$\sin { \left( \angle C \right) } =\sqrt { 1-\cos ^{ 2 }{ \left( \angle C \right) } } =\frac { \sqrt { 6479 } }{ 180 }$

Now let x=AO $\qquad \Rightarrow$ OC=25-x

$\triangle AOD$ and $\triangle COE$ are right angle //since OE and OD are radius

$\sin { \left( \angle A \right) } =\frac { radius }{ x } \Rightarrow radius=x\cdot \sin { \left( \angle A \right) }$

$\sin { \left( \angle C \right) } =\frac { radius }{ 25-x } \Rightarrow radius=\left( 25-x \right) \sin { \left( \angle C \right) }$

$\Rightarrow x\cdot \sin { \left( \angle A \right) } =\left( 25-x \right) \sin { \left( \angle C \right) }$

$x=\frac { 25\sin { \left( \angle C \right) } }{ \sin { \left( \angle C \right) } +\sin { \left( \angle A \right) } } =\frac { 25\frac { \sqrt { 6479 } }{ 180 } }{ \frac { \sqrt { 6479 } }{ 180 } +\frac { \sqrt { 6479 } }{ 120 } }$

multiply denominator and numerator by $\frac { 180 }{ \sqrt { 6479 } }$

$\Rightarrow x=\frac { 25 }{ \frac { 5 }{ 2 } } =10$