The image shows a geometric problem where:

• $DE \parallel QR$,
• $AP$ and $BP$ are angle bisectors of $\angle EAB$ and $\angle RBA$, respectively,
• We are required to find $\angle APB$.

Step-by-step Solution:

1. Since $DE \parallel QR$, and lines $AB$ intersect these parallel lines, we can apply properties of transversals and parallel lines to understand the angle relationships.

2. Let $\angle EAB = \theta_1$ and $\angle RBA = \theta_2$.

3. Since $AP$ and $BP$ are the bisectors of $\angle EAB$ and $\angle RBA$, the angles at $P$ are half of $\theta_1$ and $\theta_2$ respectively: $\angle PAB = \frac{\theta_1}{2}, \quad \angle PBA = \frac{\theta_2}{2}$

4. Using the exterior angle theorem (as lines $DE \parallel QR$), we know that: $\theta_1 + \theta_2 = 180^\circ$

5. Now, we can calculate $\angle APB$ by using the angle sum property in the triangle: $\angle APB = 180^\circ - \left( \frac{\theta_1}{2} + \frac{\theta_2}{2} \right)$ Substituting $\theta_1 + \theta_2 = 180^\circ$: $\angle APB = 180^\circ - \frac{180^\circ}{2} = 90^\circ$

Thus, $\angle APB = 90^\circ$.

Let me know if you would like further clarifications, and here are some additional questions for practice:

1. If $\theta_1 = 120^\circ$ and $\theta_2 = 60^\circ$, what are the individual angles at $P$?
2. What other geometric properties could we derive from the parallel lines in the diagram?
3. If we change the orientation of the lines $DE$ and $QR$, will the result for $\angle APB$ change?
4. How would the solution change if $AP$ and $BP$ were not bisectors?
5. Can we generalize this method for other configurations of parallel lines and angle bisectors?

Tip: Always use the parallel line theorem when you have parallel lines intersected by a transversal—it simplifies angle calculations!