The points $A,B,C,D$ are collinear. The point $P$ sits off the line, and $\angle{APB}=\angle{CPD}=\theta.$
I'd like to show that if the points $P,A,D$ are fixed, the ratio $\dfrac{AB\cdot AC}{DC\cdot DB}$ is independent of $\theta$.
I noticed that this ratio is equal to $(PA/PD)^2$, so it would suffice to show that these quantities are the same.
I'd be especially interested in knowing whether this result is well known or has a name. This problem came up when I was investigating a special case of Carnot's Theorem for Conics.
This is a well-known result known as Steiner's ratio theorem. The proof uses essentially just the law of sines, but I'll reproduce it here.
Lemma: (sometimes known as the ratio theorem). For $\triangle ABC$ with $D$ on $BC$, we have $$\frac{BD}{DC}=\frac{AB}{AC}\frac{\sin(\angle BAD)}{\sin(\angle DAC)}.$$ Proof of lemma: By law of sines, \begin{align*} \frac{BD}{DC}&=\frac{BD}{AB}\frac{AC}{DC}\frac{AB}{AC}\\ &=\frac{\sin(\angle BAD)}{\sin(\angle BDA)}\frac{\sin(\angle ADC)}{\sin(\angle DAC)}\frac{AB}{AC}\\ &=\frac{AB}{AC}\frac{\sin(\angle BAD)}{\sin(\angle DAC)} \end{align*} as claimed.
Then using the lemma, we have $$\frac{AB}{BD}\frac{AC}{CD}=\left(\frac{PA}{PD}\right)^2\frac{\sin(\theta)}{\sin(\angle APD-\theta)}\frac{\sin(\angle APD-\theta)}{\sin(\theta)}=\left(\frac{PA}{PD}\right)^2.$$