Proof of the intersecting secants theorem by the power of a point

Proof

We show that each product equals the absolute value of $d(P, O)^2 - r^2$.

1. The points are concyclic in $\mathbb{R}^2$. So there are a point $O$ and a number $r \geq 0$ with $d(X, O) = r$ for $X = A, B, C, D$, where $d$ is the Euclidean distance.
2. For a unit vector $u$, put $q(t) = t^2 + 2t\, u \cdot (P - O) + k$ with $k = \norm{P - O}^2 - r^2$. With the dot product, $\norm{P + tu - O}^2 - r^2 = q(t)$.
3. Claim: let $X \neq Y$ be points at distance $r$ from $O$, and let $P$ be on the line $XY$. Then $d(P, X)\, d(P, Y) = \abs{k}$.
4. Proof of the claim: let $u = (Y - X)/\norm{Y - X}$. The line $XY$ is $\set{P + tu : t \in \mathbb{R}}$, since it contains $P$ and has the direction $u$. So $X = P + t_1 u$ and $Y = P + t_2 u$, with $t_1 \neq t_2$ since $X \neq Y$.
5. By step 2, $q(t_1) = q(t_2) = 0$. Write $q(t) = t^2 + pt + k$.
6. Subtracting, $0 = q(t_1) - q(t_2) = (t_1 - t_2)(t_1 + t_2 + p)$. Since $t_1 \neq t_2$, we get $p = -(t_1 + t_2)$.
7. Then $0 = q(t_1) = t_1^2 - (t_1 + t_2)t_1 + k = k - t_1 t_2$. So $t_1 t_2 = k$.
8. Since $u$ is a unit vector, $d(P, X) = \abs{t_1}$ and $d(P, Y) = \abs{t_2}$. So $d(P, X)\, d(P, Y) = \abs{t_1 t_2} = \abs{k}$, which proves the claim.
9. Apply the claim to $X = A$, $Y = D$, which is allowed since $A \neq D$ and $P$ is on the line $AD$. This gives $d(P, A)\, d(P, D) = \abs{k}$.
10. Apply it to $X = B$, $Y = C$, since $B \neq C$ and $P$ is on the line $BC$. This gives $d(P, B)\, d(P, C) = \abs{k}$.
11. Hence $d(P, A)\, d(P, D) = d(P, B)\, d(P, C)$. Since $P$ is outside the circle, $k > 0$, and both products equal the power $k$ of $P$.