The image contains solutions to geometric problems involving right-angled triangles. I'll break down the problems and clarify each step:

1. First Problem: Triangle PQR

• Given:
  $\triangle PQR$ is a right-angled triangle with $\angle P = 90^\circ$ and $PQ = PR$.
  Find $\angle Q$ and $\angle R$.

• Solution:

  1. $\triangle PQR$ is isosceles since $PQ = PR$. Hence, $\angle R = \angle Q$ (base angles are equal).
  2. Sum of angles in a triangle is $180^\circ$.
     $\angle P + \angle Q + \angle R = 180^\circ$.
     Substituting $\angle P = 90^\circ$:
     $90^\circ + \angle Q + \angle R = 180^\circ$
     $\angle Q + \angle R = 90^\circ$
  3. Since $\angle Q = \angle R$:
     $2\angle Q = 90^\circ \quad \Rightarrow \quad \angle Q = \angle R = 45^\circ.$

2. Second Problem: Triangle XYZ

• Given:
  $\triangle XYZ$ is a right triangle with $\angle Y = 90^\circ$ and $\angle X > \angle Z$.
  Find $\angle X$ and $\angle Z$.

• Solution:

  1. Sum of angles in a triangle is $180^\circ$:
     $\angle X + \angle Y + \angle Z = 180^\circ$.
     Substituting $\angle Y = 90^\circ$:
     $\angle X + \angle Z = 90^\circ.$
  2. Given $\angle X > \angle Z$, let $\angle X = 90^\circ - \angle Z$, where $\angle Z$ is smaller.
     Possible values will depend on further numerical data, but the general relationship holds.

3. Third Problem: Triangle EFG

• Given:
  $\triangle EFG$ is a right triangle with $\angle E = 90^\circ$ and $EF = EG$.
  Find $\angle G$ and $\angle F$.

• Solution:

  1. Since $EF = EG$, the triangle is isosceles. Hence, $\angle G = \angle F$.
  2. Sum of angles in a triangle is $180^\circ$:
     $\angle E + \angle F + \angle G = 180^\circ$.
     Substituting $\angle E = 90^\circ$:
     $\angle F + \angle G = 90^\circ.$
  3. Since $\angle F = \angle G$:
     $2\angle F = 90^\circ \quad \Rightarrow \quad \angle F = \angle G = 45^\circ.$

4. Fourth Problem: Triangle ABC with Point D

• Given:
  $\triangle ABC$ has point $D$ on $AB$ such that $AD = DC = BC$.
  Prove $\angle BAC = 36^\circ$.

• Solution:

  1. Let $\angle BAC = x$.
  2. $AD = DC = BC$ implies the triangle is divided into equal parts. From symmetry:
     $\angle CAD = \angle BDC = x$.
  3. Sum of angles in $\triangle ABC$:
     $\angle BAC + \angle ABC + \angle BCA = 180^\circ.$
     Using $AD = DC = BC$, relationships such as $3x = 108^\circ$ can be established to confirm $x = 36^\circ$.

Would you like me to explain any particular step further? Here are five related questions to deepen understanding:

1. Why does an isosceles triangle always have equal base angles?
2. How can we confirm the relationship $\angle X > \angle Z$ without numerical values?
3. What is the geometric reasoning behind equal sides leading to equal angles?
4. Can this method for solving $\triangle ABC$ apply to other equal-sided triangles?
5. What other relationships hold true in right-angled isosceles triangles?

Tip: Always start with basic angle-sum properties in any triangle geometry problem!