1. **Problem statement:** We have triangle ABC with point D on segment AC such that BD is perpendicular to AC and BD bisects the angle at B. Given lengths are $AD=5$ and $BD=12$. We need to find the length of $BC$ rounded to the nearest tenth. 2. **Key concepts:** Since BD bisects the angle at B and is perpendicular to AC, triangle ABD and triangle CBD are right triangles sharing BD as a height. 3. **Using the Pythagorean theorem:** In right triangle ABD, $$AB=\sqrt{AD^2+BD^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13.$$ 4. Since BD bisects the angle at B, triangles ABD and CBD are congruent, so $AD=DC=5$. 5. Therefore, $AC=AD+DC=5+5=10$. 6. In right triangle BDC, $$BC=\sqrt{DC^2+BD^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13.$$ 7. **Final answer:** The length of $BC$ is $13.0$ units.