Question

In chords AB and CD intersect at P. Prove that PB.PB= CP. PD

Answer 1

Proof
CASE 1 (The intersection point is inside the circle.)
Consider triangles APC and BPD.
∠APC = ∠BPD (vertically opposite)
∠CDB = ∠CAB (angles in the same segment)
∠ACD = ∠DBA (angles in the same segment)
Therefore triangle CAP is similar to triangle BDP.
Therefore
AP
PD = CP
PB
and AP · PB = CP · PD, which can be written PA · PB = PC · PD
CASE 2 (The intersection point is outside the circle.)
Show triangle APD is similar to triangle CPB
Hence
AP
CP = PD
PB
i.e. AP · PB = PD.CP
which can be written PA · PB = PC · PD