- Let $l$ be a line and $A$ be a fixed point.
Draw a line through $A$ meeting $l$ at $B$.
Take the point $C$ on the half-line $BA$ such that $BC$ equals a given constant.
Draw the locus of $C$ (called conchoid).
Let the tangent of the locus at $C$ intersect $l$ at the point $D$.
I attempted to prove$$\tag1\vec{BD}\cdot\vec{CD}=\frac{ BD^2 AB+CD^2 AC}{AB+AC}$$ where $\vec{BD}\cdot\vec{CD}=BD\cdot CD\cos{\angle BDC}$. - Let $l$ be a line and $A,P$ be fixed points.
Draw a line through $A$ meeting $l$ at $B$.
Take the point $C$ on the half-line $BA$ such that $BC=BP$. Draw the locus of $C$.
Let the tangent of the locus at $C$ intersect $l$ at the point $D$.
I attempted to prove$$\tag2\vec{BD}\cdot\vec{CD}=\frac{BD^2 AB+CD^2 AC - AB\cdot BD\cdot BP \cos{\angle DBP}}{AB+AC}$$
For (1):
First, I write inner products in terms of length
$$\vec{BD}\cdot\vec{CD}=\frac{BD^2+CD^2-BC^2}{2}$$
(1) becomes
$$\frac{BD^2+CD^2-BC^2}{2}=\frac{ BD^2 AB+CD^2 AC}{AB+AC}$$
Multiplying by $2(AB+AC)$ and rearranging
$$(AB-AC)(CD^2-BD^2)=BC^2(AB+AC)$$
Dividing by $(AB-AC)=BC$
$$CD^2-BD^2=BC(AB+AC)$$
Let $A'$ be symmetric point of $A$ about the midpoint of $BC$.
Using $BC=CA'-BA'$ and $AB+AC=CA'+BA'$, the equation becomes
$$CD^2-BD^2=CA'^2-BA'^2$$
which is equivalent to $A'D\perp AB$,
which is equivalent to
$$A'C\tan\angle A'CD=A'B\tan\angle A'BD$$
which is equivalent to
$$\frac{A'C}{\tan\angle A'BD}=\frac{A'B}{\tan\angle A'CD}$$
using $A'C=AB,A'B=AC,\tan\angle A'BD=\tan\angle ABD,\tan\angle A'CD=\tan\angle ACD,$ the equation becomes
$$\frac{AB}{\tan\angle ABD}=\frac{AC}{\tan\angle ACD}\tag3$$
To prove (3), consider two points $B,B'$ on the line $l$ and take $C,C'$ on the line $AB,AB'$ such that $B'C'=BC$.
Keep $B,C$ fixed and move $B'$ towards $B$, so $C'$ moves towards $C$. The tangent of the locus at $C$ is the limit of the line $CC'$.
Let $D$ be intersection of $CC'$ and $l$, and $E,F$ be projection of $B',C'$ on the line $AB$.
Since $\angle ABD=\pi-\angle ABB',\angle ACD=\pi-\angle ACC'$ and ${B'E\over C'F}={AB'\over AC'}$ we have
\begin{align}
{BE\over CF}&={B'E/\tan\angle ABB'\over C'F/\tan\angle ACC'}
\\&={AB'/\tan\angle ABD\over AC'/\tan\angle ACD}\to{AB/\tan\angle ABD\over AC/\tan\angle ACD}\quad\text{as }B'\to B
\end{align}
But since $BC=B'C'$ we have ${BE\over CF}\to1$, then ${AB/\tan\angle ABD\over AC/\tan\angle ACD}=1$, so we get (3), so we proved (1).
For (2):
I attempted to rewrite (2) into a simple form like (3) but didn't succeed.
How to prove (2)?
Interestingly, (2) is the same as (1) if $PB\perp l$ (Then $\cos\angle DBP=0$ and the last term vanishes).