All-Soviet-Union Mathematical Olympiad

Solution:

Compare the triangles $A'B'A$ and $ADB$. The base of $A'B'A$ can be taken as $A'A$, which is the same length as $AD$. The height of $A'B'A$ is $AB'$ times $\sin B'A'A'$, which is twice $AB$ times $\sin BAD$. So $\text{area } A'B'A = 2 \times \text{area } ADB$.

Similarly, $\text{area } B'C'B = 2 \times \text{area } BAC$, $\text{area } C'D'C = 2 \times \text{area } CBD$, and $\text{area } D'A'D = 2 \times \text{area } DCA$.

So adding, the $\text{area } A'B'A + \text{area } C'D'C = 2 \times \text{area } ABCD$, and $\text{area } B'C'B + \text{area } D'A'D = 2 \times \text{area } ABCD$.

But $ABCD = A'B'A + B'C'B + C'D'C + D'A'D + ABCD$. Hence result.