If

$$
a: b=c: d
$$

and

$$
a_{1}: b_{1}=c_{1}: d_{1}
$$

then what is the condition for

$$
\left(a+a_{1}\right):\left(b+b_{1}\right)=\left(c+c_{1}\right):\left(d+d_{1}\right)
$$
From the third proportion, we get:

$$
a d + a d_{1} + a_{1} d + a_{1} d_{1} = b c + b_{1} c + b c_{1} + b_{1} c_{1}
$$

But from the first and second proportions:

$$
a d = b c, a_{1} d_{1} = b_{1} c_{1},
$$

taking into account (1), it follows from (1) that:

$$
a d_{1} + a_{1} d = b c_{1} + b_{1} c
$$

(Balazs Toth, Eger.)

The problem was also solved by: Baumgarten I., Freund E., Füstös P., Harsányi Z., Jánosy Gy., Makó E., Reich Z., Schlesinger Gy., Schuster Gy.