## Task 4 - 130814

In a trapezoid $A B C D$ with $A B \| C D$ and $\overline{A B}>\overline{C D}$, let $\overline{A B}=a, \overline{C D}=c$ and the distance $h$ between the parallel sides be given. The diagonals $A C$ and $B D$ intersect the midline $F E$ of the trapezoid at $H$ and $G$, respectively.

Determine the area $A_{T}$ of the trapezoid $A B G H$!
The area $A_{T}$ of trapezoid $A B G H$ is equal to the difference between the area of trapezoid $A B E F$ and the sum of the areas of triangles $A H F$ and $B E G$.

Now, the area of triangle $C D A$ is equal to that of triangle $C D B$, namely $\frac{c \cdot h}{2}$. The area of triangle $C D H$ is equal to that of triangle $C D G$, which is half the size of triangle $C D A$, thus equal to $\frac{c \cdot h}{4}$.

Therefore, the area of each of the triangles $D A H$ and $B C G$ is also equal to $\frac{c \cdot h}{4}$.

Finally, the area of triangle $A H F$ and similarly that of triangle $B E G$ (due to $A F = F D$ and $B E = E C$) is half the size of triangle $D A H$, thus equal to $\frac{c \cdot h}{8}$.

The area of trapezoid $A B E F$ is

$$
\frac{1}{2}\left(\frac{a+c}{2}+a\right) \cdot \frac{h}{2}=\frac{h}{8}(3 a+c)
$$

Consequently, the area of trapezoid $A B G H$ is

$$
A_{T}=\frac{h}{8}(3 a+c)-2 \cdot \frac{c \cdot h}{8}=\frac{h}{8}(3 a-c)
$$

Solutions of the 1st Round 1973 adopted from [5]

### 5.15.2 2nd Round 1973, Class 8