Two chords AB and CD of a circle intersect at a point P outside the circle. Prove that

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

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

(a). To Prove: ∆PAC ∼ ∆PDB

Proof: We know

∠ABD + ∠ACD = 180° [∵ opposite angles of cyclic quadrilateral are supplementary] …(i)

∠PCA + ∠ACD = 180° [∵ they are linear pair angle] …(ii)

Comparing equations (i) & (ii), we get

∠ABD + ∠ACD = ∠PCA + ∠ACD

⇒ ∠ABD = ∠PCA

Also, ∠APC = ∠BPD [∵ they are common angles]

Thus, by AA-similarity criteria, ∆PAC ∼ ∆PDB

Hence, proved.

(b). To Prove: PA × PB = PC × PD

Proof: We have already proved that, ∆PAC ∼ ∆PDB

Thus the ratios can be written as,

By cross-multiplication, we get

PA × PB = PC × PD

Hence, proved.