Question 8

We will first prove that the inner figure is a parallelogram. Then we will compare its area to the outer 4-gon.

Step 1 — Proving PQRS is a parallelogram

Let ABCD be our 4-gon. P, Q, R, S are midpoints. They are on sides AB, BC, CD, DA respectively. We join P, Q, R, S in order. This forms the figure PQRS. Let's draw diagonal AC. In $\triangle ABC$, P and Q are midpoints. By the Midpoint Theorem, PQ is parallel to AC. Also, PQ is half of AC. In $\triangle ADC$, S and R are midpoints. By the Midpoint Theorem, SR is parallel to AC. Also, SR is half of AC. So, PQ is parallel to SR. And PQ equals SR. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. Thus, PQRS is a parallelogram.

Step 2 — Comparing areas

We know PQRS is a parallelogram. Let's recall a property of triangles. When we join two midpoints of its sides, a smaller triangle forms. Its area is $\frac{1}{4}$ of the original triangle's area. Let's consider diagonal AC. In $\triangle ABC$, P and Q are midpoints. So, Area($\triangle PBQ$) is $\frac{1}{4}$ of Area($\triangle ABC$). In $\triangle ADC$, S and R are midpoints. So, Area($\triangle SDR$) is $\frac{1}{4}$ of Area($\triangle ADC$). Let's add these two areas.

$\text{Area}(\triangle PBQ) + \text{Area}(\triangle SDR) = \frac{1}{4} \text{Area}(\triangle ABC) + \frac{1}{4} \text{Area}(\triangle ADC)$

$= \frac{1}{4} (\text{Area}(\triangle ABC) + \text{Area}(\triangle ADC))$

$= \frac{1}{4} \text{Area}(\text{ABCD})$

Now, let's draw diagonal BD. In $\triangle ABD$, P and S are midpoints. So, Area($\triangle APS$) is $\frac{1}{4}$ of Area($\triangle ABD$). In $\triangle BCD$, Q and R are midpoints. So, Area($\triangle CQR$) is $\frac{1}{4}$ of Area($\triangle BCD$). Let's add these two areas.

$\text{Area}(\triangle APS) + \text{Area}(\triangle CQR) = \frac{1}{4} \text{Area}(\triangle ABD) + \frac{1}{4} \text{Area}(\triangle BCD)$

$= \frac{1}{4} (\text{Area}(\triangle ABD) + \text{Area}(\triangle BCD))$

$= \frac{1}{4} \text{Area}(\text{ABCD})$

The total area of the four corner triangles is their sum.

$\text{Total corner area} = \text{Area}(\triangle PBQ) + \text{Area}(\triangle SDR) + \text{Area}(\triangle APS) + \text{Area}(\triangle CQR)$

$= \frac{1}{4} \text{Area}(\text{ABCD}) + \frac{1}{4} \text{Area}(\text{ABCD})$

$= \frac{1}{2} \text{Area}(\text{ABCD})$

The area of parallelogram PQRS is the 4-gon area minus the corner triangles.

$\text{Area}(\text{PQRS}) = \text{Area}(\text{ABCD}) - \text{Total corner area}$

$= \text{Area}(\text{ABCD}) - \frac{1}{2} \text{Area}(\text{ABCD})$

$\boxed{\text{Area}(\text{parallelogram PQRS}) = \frac{1}{2} \times \text{Area}(\text{4-gon ABCD})}$