Solution:

Given the approximately equal fractions
$$\frac{a}{b} \text{ and } \frac{c}{d}$$
we multiply the four terms by $a+b+c+d$ to get
$$\frac{a^{2}+a b+a c+a d}{a b+b^{2}+b c+b d} \quad \text{ and } \quad \frac{a c+b c+c^{2}+c d}{a d+b d+c d+d^{2}}$$
We then adjust each member by 1 by changing the $a d$ terms to $b c$ and vice versa. The resulting fractions can be factorized:
$$\begin{aligned} & \frac{a^{2}+a b+a c+b c}{a b+b^{2}+a d+b d} \quad \text{ and } \quad \frac{a c+a d+c^{2}+c d}{b c+b d+c d+d^{2}} \\ & \frac{(a+c)(a+b)}{(b+d)(a+b)} \quad \text{ and } \quad \frac{(a+c)(c+d)}{(b+d)(c+d)} \text{. } \end{aligned}$$
The last pair of fractions are visibly equal.