I'm trying to understand the proof of the reciprocal Pythagorean theorem, but it seems to depend on the fact that $ab = cd$ in the picture above, but I cannot see why. Assuming the result, I can then apply Pythagoras getting $a^2 + b^2 = c^2$ which implies $1/b^2 + 1/a^2 = c^2/(a^2b^2)$ which implies $1/b^2 + 1/a^2 = c^2/(c^2d^2)$ which implies $1/b^2 + 1/a^2 = c^2/(c^2d^2)$ ending up with $1/b^2 + 1/a^2 = 1/d^2$, as desired. (So I'm missing only why $ab = cd$.)
Let the angle between $a$ and $c$ be $\alpha$. Note that $\sin(\alpha)=\cos(90^\circ-\alpha)$.
Therefore $\sin(\alpha)=\frac{d}{a}$, so $a=\frac{d}{sin(\alpha)}$.
Similarly, $\cos(90^\circ-\alpha)=\sin(\alpha)=\frac{b}{c}$, so $b=c\sin(\alpha)$.
Therefore $ab=cd$.