In the figure $(i i)$ given below, chords $B A$ and $D C$ of a circle meet at $P$ . Prove that
i) $∠ P A C = ∠ P D B$
ii) $P A \times P B = P C \times P D$ .

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Solution

Solution:
Construction: Join AC and BD
Proof:
(i) We know that in a cyclic quadrilateral, the exterior angle is equal to opposite interior angle.
Therefore, $∠ P A C = ∠ P D B$ ……….(i) Proved
Similarly, $∠ P C A = ∠ P B D$ …………(ii)
In view of (i) and (ii)
$Δ P A C \sim Δ P A B$ (AA test of similarity)
$∴ \frac{P A}{P D} = \frac{P C}{P B}$ (since $Δ P A C \sim Δ P A B$ ) & corresponding sides of similar triangles are proportional
$∴ P A \times P B = P C \times P D$
Hence proved.

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In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that i) ∠PAC=∠PDBii) PA×PB=PC×PD.

In the figure $(i i)$ given below, chords $B A$ and $D C$ of a circle meet at $P$ . Prove that
i) $∠ P A C = ∠ P D B$
ii) $P A \times P B = P C \times P D$ .