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Question
Two chords AB and CD of a circle intersect at a point P outside the circle.
Prove that: (i) Δ PAC ~ Δ PDB (ii) PA. PB = PC.PD
Solution
Given : AB and CD are two chords
To Prove:
(a) Δ PAC - Δ PDB
(b) PA. PB = PC.PD
Proof: ∠𝐴𝐵𝐷 + ∠𝐴𝐶𝐷 = 180° …(1) (Opposite angles of a cyclic quadrilateral are
supplementary)
∠𝑃𝐶𝐴 + ∠𝐴𝐶𝐷 = 180° …(2) (Linear Pair Angles )
Using (1) and (2), we get
∠𝐴𝐵𝐷 = ∠𝑃𝐶𝐴
∠𝐴 = ∠𝐴 (Common)
By AA similarity-criterion Δ PAC - Δ PDB
When two triangles are similar, then the rations of the lengths of their corresponding sides are proportional
`∴ (PA)/(PD)=(PC)/(PB)`
⟹ PA.PB = PC.PD
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