To solve this problem, let's break down the information provided:

1. Angle BPD: Since $\angle BPD$ is an inscribed angle that intercepts arc $BD$ of $70^\circ$, we can use the inscribed angle theorem. This theorem states that an inscribed angle is half of the measure of the intercepted arc. Therefore: $\angle BPD = \frac{1}{2} \times m \overset{\frown}{BD} = \frac{1}{2} \times 70^\circ = 35^\circ$

2. Sum of $m \overset{\frown}{BC} + m \overset{\frown}{AD}$: The circle has a total arc measure of $360^\circ$. Given that $m \overset{\frown}{BD} = 70^\circ$ and $m \overset{\frown}{CA} = 170^\circ$, we can find the remaining arcs $m \overset{\frown}{BC}$ and $m \overset{\frown}{AD}$ by subtracting from $360^\circ$: $m \overset{\frown}{BC} + m \overset{\frown}{AD} = 360^\circ - (m \overset{\frown}{BD} + m \overset{\frown}{CA}) = 360^\circ - (70^\circ + 170^\circ) = 120^\circ$

Final Answers:

• $\angle BPD = 35^\circ$
• $m \overset{\frown}{BC} + m \overset{\frown}{AD} = 120^\circ$

Would you like more details on any of these steps?

Here are 5 related questions to deepen understanding:

1. What is the inscribed angle theorem, and why does it apply here?
2. How do we find other arc measures in a circle if only a few arcs are given?
3. Can you explain how central angles differ from inscribed angles?
4. What if $m \overset{\frown}{CA}$ was different; how would it change $m \overset{\frown}{BC} + m \overset{\frown}{AD}$?
5. How can we use symmetry in circles to simplify problems with arcs and angles?

Tip: In circle problems, identifying intercepted arcs can simplify finding angle measures.