4 Questions around this concept.
Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 2 cm, BP = 6 cm, and CP = 2.5 cm. Then PD is:

Result 1: If two chords of a circle intersect each other, then the products of the lengths of their segments are equal.
Let us prove this
Case 1: Let the two chords intersect internally (inside the circle).

Let AB and CD are two chords intersecting at point P in the circle, and we have to prove (PA) (PB) = (PC) (PD).
Join AC and BD.
In triangles APC and PDB,
∠APC = ∠DPB (Vertically opposite angels)
∠CAP = ∠CDB (Angles made by arc BC in the same segment)
ΔAPC is similar to ΔDPB. (by the AA Similarity Property)
Case 2: Let the two chords intersect externally (outside the circle).

Let two chords BA and CD intersect at point P which lies outside the circle, and we have to prove (PA) (PB) = (PC) (PD).
Join AC and BD.
In triangles PAC and PDB,
∠PAC = ∠PDB
and ∠PCA = ∠PBD
(An external angle of a cyclic quadrilateral is equal to the interior angle at the opposite vertex)
ΔPAC is similar to ΔPDB. (by the AA Similarity Property)
Result 2: The converse of the above Result is also true, i.e., if two line segments AB and CD intersect at P and (PA)(PB) = (PC)(PD), then the four points are concyclic.
Result 3: If one of the secants (say PCD) is rotated around P so that it becomes a tangent, i.e., points C and D say at T, coincide. We get the following result:

If PAB is a secant to a circle intersecting the circle at A and B and PT is the tangent drawn from P to the circle, then PA · PB = PT2
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