Intersecting secants theorem

Statement

Let $A$, $B$, $C$ and $D$ be concyclic points of the Euclidean plane $\mathbb{R}^2$, with $A \neq D$ and $B \neq C$. Let $d$ be the Euclidean distance. Suppose that the lines $AD$ and $BC$ meet at a point $P$ outside the circle. Then $$d(P, A)\, d(P, D) = d(P, B)\, d(P, C).$$

Remarks

Both products are equal to the power $d(P, O)^2 - r^2$ of $P$, where $O$ is the centre and $r$ the radius. A proof: the triangles $PAC$ and $PBD$ are similar, because they have the same angle at $P$ and equal inscribed angles at $C$ and $D$. If the second line becomes tangent, then $B = C$, and the statement becomes the tangent–secant theorem.

Explanations

L1 human

From a point outside a circle, draw two lines that cut the circle. On each line, multiply the distances from the point to the two crossing points. The two products are equal.