1. **Problem statement:** Find the length of side $BC$ in triangle $ABC$ where $AB=17$, $AC=8$, and there is a right angle at the foot of the altitude from $C$ to $AB$. 2. **Key concept:** The altitude from $C$ to $AB$ creates two right triangles within $ABC$. We can use the Pythagorean theorem and properties of right triangles to find $BC$. 3. **Step 1:** Let the foot of the altitude from $C$ to $AB$ be $D$. Then $CD$ is perpendicular to $AB$. 4. **Step 2:** Since $CD$ is an altitude, triangles $ADC$ and $BDC$ are right triangles. 5. **Step 3:** Use the Pythagorean theorem in triangle $ADC$: $$AC^2 = AD^2 + CD^2$$ $$8^2 = AD^2 + CD^2$$ $$64 = AD^2 + CD^2$$ 6. **Step 4:** Use the Pythagorean theorem in triangle $BDC$: $$BC^2 = BD^2 + CD^2$$ 7. **Step 5:** Since $D$ lies on $AB$, $AD + BD = AB = 17$. 8. **Step 6:** Express $BD$ as $17 - AD$ and substitute into the equation for $BC^2$: $$BC^2 = (17 - AD)^2 + CD^2$$ 9. **Step 7:** From step 5, $CD^2 = 64 - AD^2$. Substitute into $BC^2$: $$BC^2 = (17 - AD)^2 + 64 - AD^2$$ 10. **Step 8:** Expand and simplify: $$BC^2 = (289 - 34AD + AD^2) + 64 - AD^2 = 289 - 34AD + 64 = 353 - 34AD$$ 11. **Step 9:** To find $AD$, use the fact that $CD$ is the altitude to hypotenuse $AB$ in right triangle $ABC$. The altitude to the hypotenuse satisfies: $$CD = \frac{AC \times BC}{AB}$$ 12. **Step 10:** Using the Pythagorean theorem for triangle $ABC$: $$AB^2 = AC^2 + BC^2$$ $$17^2 = 8^2 + BC^2$$ $$289 = 64 + BC^2$$ $$BC^2 = 225$$ $$BC = 15$$ 13. **Final answer:** The length of $BC$ is $15$ units.