The problem you've provided asks to show that:

$4 \times \text{Area}(\triangle PQZ) = \text{Area}(\triangle XYZ)$

where $P$ and $Q$ are the midpoints of $YZ$ and $XZ$ respectively.

Proof:

1. Identify the Triangle and Midpoints:

   • $P$ is the midpoint of $YZ$.
   • $Q$ is the midpoint of $XZ$.

2. Apply the Midpoint Theorem:

   • According to the midpoint theorem, the line segment joining two midpoints of a triangle is parallel to the third side and half its length. Therefore, $PQ$ is parallel to $XY$ and $PQ = \frac{1}{2}XY$.

3. Area Calculation:

   • The area of a triangle is given by $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.
   • Since $PQ$ is parallel to $XY$ and half its length, the height from $Z$ to $PQ$ in $\triangle PQZ$ is half the height from $Z$ to $XY$ in $\triangle XYZ$.
   • Therefore, the area of $\triangle PQZ$ is $\frac{1}{2} \times \frac{1}{2} \times$ area of $\triangle XYZ$ $= \frac{1}{4} \times$ area of $\triangle XYZ$.

4. Conclusion:

   • Hence, $4 \times \text{Area}(\triangle PQZ) = \text{Area}(\triangle XYZ)$.

This completes the proof.

Do you have any questions or would you like further details?

Here are 5 related questions:

1. How does the midpoint theorem apply to different types of triangles, such as scalene or equilateral?
2. Can this proof be generalized to any polygon, or is it specific to triangles?
3. What would happen if $P$ and $Q$ were not midpoints? How would the areas compare?
4. How does the orientation of triangle $XYZ$ (acute, obtuse) affect the validity of this area relationship?
5. How can the centroid of a triangle be used in similar area calculations?

Tip: When solving problems involving midpoints and areas, visualizing the triangle and applying geometric properties like the midpoint theorem can simplify complex proofs.