The problem in the image provides two options. Let's solve both:

Problem 1:

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that AD bisects ∠A.

Solution:

1. Since $AB = AC$, triangle ABC is isosceles, meaning that the angles opposite these sides, i.e., $\angle B = \angle C$.

2. The altitude $AD$ is drawn from vertex $A$ to the base $BC$, making $AD$ perpendicular to $BC$. Therefore, $\angle ADB = \angle ADC = 90^\circ$.

3. In triangle $ABD$ and triangle $ACD$:

   • $AB = AC$ (given).
   • $AD = AD$ (common side).
   • $\angle ADB = \angle ADC = 90^\circ$ (altitude property).

   Therefore, by the RHS congruence criterion, $\triangle ABD \cong \triangle ACD$.

4. Since the two triangles are congruent, $\angle BAD = \angle CAD$. This shows that $AD$ bisects $\angle A$.

Problem 2:

In the figure below, the diagonal PR of quadrilateral PQRS bisects $\angle QPS$ and $\angle QRS$. Prove that QR = SR.

Solution:

1. The diagonals $PR$ bisect $\angle QPS$ and $\angle QRS$, so:

   • $\angle QPR = \angle SPR$ (since $PR$ bisects $\angle QPS$).
   • $\angle QRP = \angle SRP$ (since $PR$ bisects $\angle QRS$).

2. In triangles $QPR$ and $SPR$:

   • $\angle QPR = \angle SPR$ (given).
   • $\angle QRP = \angle SRP$ (given).
   • $PR = PR$ (common side).

   Therefore, by the ASA congruence criterion, $\triangle QPR \cong \triangle SPR$.

3. Since the triangles are congruent, we conclude that $QR = SR$.

Would you like further details or explanations? Here are some related questions for deeper understanding:

1. What are the properties of an isosceles triangle?
2. How does congruence work in triangle proofs?
3. What is the significance of the altitude in an isosceles triangle?
4. Can a diagonal in a quadrilateral always bisect opposite angles?
5. What are the different criteria for triangle congruence (SSS, SAS, ASA, RHS)?

Tip: In geometry, always look for congruent triangles to prove equal lengths or bisected angles!