Let $a_1+d_1=a_2=a_3-d_1$ and $b_1+d_2=b_2=b_3-d_2$. Then, $a_2b_2=(a_2-d_1)(b_2-d_2)=(a_2+d_1)(b_2+d_2)$. This means that $d_1b_2+d_2a_2=d_1d_2$ and $d_1b_2+d_2a_2=-d_1d_2$. This means that $d_1d_2=0$. Since $d_2\neq0$, we must have $d_1=0$, so $d_2a_2=0$, which is a contradiction.

DottedCaculator wrote: Let $a_1+d_1=a_2=a_3-d_1$ and $b_1+d_2=b_2=b_3-d_2$. Then, $a_2b_2=(a_2-d_1)(b_2-d_2)=(a_2+d_1)(b_2+d_2)$. This means that $d_1b_2+d_2a_2=d_1d_2$ and $d_1b_2+d_2a_2=-d_1d_2$. This means that $d_1d_2=0$. Since $d_2\neq0$, we must have $d_1=0$, so $d_2a_2=0$, which is a contradiction. This solution is wrong. The dimensions of the three rectangles do not need to be $a_1,b_1$; $a_2,b_2$; $a_3,b_3$ in this order. Still it's a weird problem.

Let's change $a_1,a_2,a_3$ to $a-x,a,a+x$ and $b_1,b_2,b_3$ to $b-y,b,b+y$ where $x,y\ne 0$. Let's say couple to 1 rectangle's base and altitude and size of couple be the area of respective rectangle . If $a-x$ makes a couple with $b-y$ or $b$,then see the couple which contains $b+y$. This couple's size will be larger than other one. So contracidtion. It means 1 of the couples is $(a-x,b+y)$. With the same idea we get other couples are $(a,b)$ and $(a+x,b-y)$. So $(a-x)(b+y)=ab=(a+x)(b-y)$. From $(1)=(3)$we get $ay-bx=bx-ay$ which means $ay=bx$. From $(1)=(2)$ we get $ab+ay-bx-xy=ab$ and since $ay=bx$ we get $xy=0$, which is contradiction!