## Task 4 - 230724

It is assumed that the angle bisectors of $\angle D A B$ and $\angle A B C$ of parallelogram $A B C D$ intersect at a point $E$ which lies on the segment $C D$ between $C$ and $D$. Furthermore, it is assumed that the segments $A E$ and $B E$ have lengths of $7 \mathrm{~cm}$ and $5 \mathrm{~cm}$, respectively. Determine the area of parallelogram $A B C D$ from these assumptions!
The triangle $A B E$ and the parallelogram $A B C D$ coincide in the side $A B$ and in the corresponding height.

Therefore, the area of the parallelogram $A B C D$ is twice as large as that of the triangle $A B E$.

Since in the parallelogram the sum of the sizes of adjacent angles is $180^{\circ}$, the sum of the angles $\angle E A B$ and $\angle A B E$ is $90^{\circ}$, so the triangle $A B E$ is a right triangle with $E$ as the vertex of the right angle.

Consequently, the area of the triangle $A B E$ is $\frac{1}{2} \cdot 7 \cdot 5 = 17.5$ and thus $17.5 \mathrm{~cm}^{2}$, and that of the parallelogram $A B C D$ is therefore $35 \mathrm{~cm}^{2}$.

Solutions of the 2nd Round 1983 taken from [5]

### 4.25.3 3rd Round 1983, Class 7