This theorem was proved by Shabbir Tejani, age 13, Jack Hunt School, Peterborough.

If two chords of a circle, produced if necessary, cut one another, the rectangle contained by the segments of the one chord (the product of the lengths of the segments) is equal to the rectangle contained by the segments of the other (again the product of the lengths of the segments).

Given that two chords $BA$, $DC$ of a circle, cutting at $P$; where $P$ lies outside the circle.

To prove that $PB \times PA=PD \times PC$.

Proof: In the triangles $BPC$ and $DPA$

Angle $B$ = angle $D$ (Same segment)

Angle $BPC = DPA$

Therefore, angle $BCP$ = angle $DAP$.

Therefore, triangles $BPC$ and $DPA$ are similar.

Therefore,

$$\frac{PB}{PD} = \frac{PC}{PA}$$

(corresponding sides proportional)

Therefore, $PB \times PA = PD \times PC$