In Fig. two chords $ AB$ and $ CD$of a circle intersect each other at the point $ P$ (when produced) outside the circle. Prove that:
$ \left(i\right)△PAC~△PDB\phantom{\rule{0ex}{0ex}}\left(ii\right)PA·PB=PC·PD$

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Solution

Step 1: To prove that $△ P A C ~ △ P D B$
In $△ P A C$ and $△ P D B,$
$∠ P = ∠ P$ [Common]
$∠ P A C = ∠ P D B$ [Exterior angle of cyclic quadrilateral is equal to its opposite interior angle]
Thus, $△ P A C ~ △ P D B$ , by $A A$ property of triangle.
Step 2: To prove that $P A \cdot P B = P C \cdot P D$
From above, $△ P A C ~ △ P D B$
We know corresponding sides of similar triangles are proportional,
$\frac{P A}{P D} = \frac{P C}{P B} \Rightarrow P A \cdot P B = P C \cdot P D$
Hence proved that $△ P A C ~ △ P D B$ and $P A \cdot P B = P C \cdot P D .$

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Similar questions

△ P A C \sim △ P D B

P A . P B = P C . P D

In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) ΔPAC ∼ ΔPDB

(ii) PA.PB = PC.PD

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

(1) $Δ P A C \sim Δ P D B$

(2) $P A . P B = P C . P D$ .

(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD

In a circle, two chords AB and CD intersect at a point P inside the circle. Prove that

(a) $Δ P A C \sim Δ P D B$

(b) $P A . P B = P C . P D$ .

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