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Question:

In Fig., two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that: (i) △PAC ∼ △PDB (ii) PA.PB = PC.PD

Solution:

(i) In△PACand△PDB,∠BAC=180∘−∠PAC(linear pairs)∠PDB=∠CDB=180∘−∠BAC=180∘−(180∘−∠PAC)=∠PAC)∠PAC=∠PDB... (1)∠APC=∠BPD[Common] ... (2)We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.Therefore, for cyclic quadrilateral ABCD.consider ABCDΔPACandΔPBD.∠PCA=∠PBD (3)∴By AAA-criterion of similarity,△PAC∼△DPB(ii)△PAC∼△DPBSo, the corresponding sides of the similar triangles are proportionalPAPD=PCPB⇒PA.PB=PC.PD