In a circle, two chords AB and CD intersect at a point inside the circle. Prove that
$(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD$

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Solution

Given: AB and CD are two chords
To Prove:
$(a) △ PAC ~ △ PDB (b) PA . PB = PC . PD$

Proof: In $△ PAC and △ PDB$
$∠ APC = ∠ DPB$ (Vertically Opposite angles)
$∠ CAP = ∠ BDP$ (Angles in the same segment are equal)
By AA similarity-criterion $△ PAC ~ △ PDB$
When two triangles are similar, then the ratios of the lengths of their corresponding sides are porportional.
$∴ \frac{PA}{PD} = \frac{PC}{PB} \Rightarrow PA . PB = PC . PD$

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Similar questions

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

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

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

(a) $Δ P A C \sim Δ P D B$

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