The diagram shows lines $AP$, $BP$, $PD$, and $PC$ intersecting at point $P$, with the given condition that $m \angle APD = m \angle BPC$. The problem asks which of the following statements must be true based on this setup:

1. $m \angle APB = 90^\circ$
2. $\frac{1}{2} m \angle APB + m \angle APD = 90^\circ$
3. $\frac{1}{2} m \angle APB + m \angle BPC = 90^\circ$

Solution:

• Statement I: $m \angle APB = 90^\circ$:
  This suggests that angle $APB$ is a right angle. However, there is no direct information in the diagram or the given condition to support this claim. Therefore, this statement is not necessarily true.

• Statement II: $\frac{1}{2} m \angle APB + m \angle APD = 90^\circ$:
  This combines part of angle $APB$ with angle $APD$, and we know from the problem that $m \angle APD = m \angle BPC$. However, the sum of these angles does not necessarily imply $90^\circ$, and no clear reasoning is provided to justify this relationship. So, this statement is also not necessarily true.

• Statement III: $\frac{1}{2} m \angle APB + m \angle BPC = 90^\circ$:
  Here, we're given that angle $APD = BPC$, and combining half of angle $APB$ with angle $BPC$ leads to a sum of $90^\circ$. This relationship could be true based on certain angle properties. Therefore, this statement is possibly true.

Conclusion:

The correct answer is (C) I and II only.

Would you like further details on the logic used or clarifications? Let me know if you have any questions!

Here are five follow-up questions:

1. How do you prove two angles are supplementary or complementary?
2. What is the relationship between vertical angles when two lines intersect?
3. Can you explain how to calculate the measure of an angle when given its complementary angle?
4. How do you solve geometric problems involving parallel lines and transversals?
5. What are the properties of exterior and interior angles in polygons?

Tip: When working with intersecting lines, always check for vertical angles, linear pairs, and possible right angles to simplify angle relationships.