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

(i) ∆ PAC ~ ∆ PDB

(ii) PA . PB = PC . PD.

In Figure, two chords AB and CD of a circle intersect each other at the point  P (when produced) outside the circle. Prove that: (i) ∆ PAC ~ ∆ PDB (ii) PA. PB =

Asked by Pragya Singh | 1 year ago |  131

1 Answer

Solution :-

(i) In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is 

∠PCA and ∠PBD is opposite interior angle, which are both equal.

∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

 

(ii) We have already proved above,

∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

Therefore,

\( \dfrac{AP}{DP}=\dfrac{PC}{PB}=\dfrac{CA}{BD}\)

\( \dfrac{AP}{DP}=\dfrac{PC}{PB}\)

AP. PB = PC. DP

Answered by Abhisek | 1 year ago

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