In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that

(a) ΔPAC ~ ΔPDB (b) PA • PB = PC • PD.

Given: AB and CD are chords of the circle, intersecting at point P.

(a). To Prove: ∆PAC ∼ ∆PDB

Proof: In ∆PAC and ∆PDB,

∠APC = ∠DPB [∵ they are vertically opposite angles]

∠CAP = ∠PDB [∵ angles in the same segment are equal]

Thus, by AA-similarity criteria, we can say that,

∆PAC ∼ ∆PDB

Hence, proved.

(b). To Prove: PA × PB = PC × PD

Proof: As already proved that ∆PAC ∼ ∆PDB

We can write as,

By cross-multiplying, we get

PA × PB = PC × PD

Hence, proved.