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(i) Δ PAC ~ Δ PDB
(ii) PA . PB = PC . PD
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(i) In triangle PAC and PDB
∠PAC + ∠CAB = 180o (Linear pair)
∠CAB + ∠BDC = 180O (Opposite angles of a cyclic quadrilateral are supplementary)
Hence,
∠PAC = ∠PDB
Similarly,
∠PCA = ∠PBD
Hence,
Δ PAC ~ Δ PDB
(ii) Since the two triangles are similar, so
Or,
PA * PB = PC * PD
Hence, proved
solved
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