From the second part $BD=DC=8$ and the fact that $D$ lies on $BC$, we have$BC= BD+DC=16$.
From the first and last part $BC= \frac{6}{AD} = \frac{6}{10}$.
Since neither $16$ nor $\frac{6}{10}$ is equal to $5$, $10$, $15$, $20$ or each other, there is no answer.
No, I cannot solve this.
From the second part $BD=DC=8$ and the fact that $D$ lies on $BC$, we have$BC= BD+DC=16$.
From the first and last part $BC= \frac{6}{AD} = \frac{6}{10}$.
Since neither $16$ nor $\frac{6}{10}$ is equal to $5$, $10$, $15$, $20$ or each other, there is no answer.