In Figure, two chords AB and CD intersect each other at the point P. Prove that :

(i) ∆APC ~ ∆ DPB

(ii) AP . PB = CP . DP

Asked by Pragya Singh | 1 year ago |  183

##### Solution :-

Firstly, let us join CB, in the given figure.

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

Therefore,

∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that ∆APC ∼ ∆DPB

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

$$\dfrac{AP}{DP}=\dfrac{PC}{PB}=\dfrac{CA}{BD}$$

$$\dfrac{AP}{DP}=\dfrac{PC}{PB}$$

AP. PB = PC. DP

Answered by Abhisek | 1 year ago

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