Question

In figure, two chords AB and CD intersect each other at the point P. Prove that
AP.PB=CP.DP​

Answer 1

Answer:

AP.BP=CP.DP

Step-by-step explanation:

Given:A circle with chords AB and CD intersecting at point P

To Prove:AP.PB=CP.DP

Proof,

In triangles APC and DPB,

<APC=<DPB(vertically opposite angles)

<CAP=<BDP(angles in the same segment of a circle are equal)

Therefore, triangles APC ~ DPB by AA similarity criterion.

ie,AP/DP=CP/BP(CPST)

.°. AP.BP=CP.DP

Hence Proved.

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Answer 2

Step-by-step explanation:

(i) Given : In △APC and △DPB,

∠APC=∠DPB ...[Vert. opp. ∠s]

∠CAP=∠BDP ...[Angles subtended by the same arc of a circle are equal]

∴ By AA-condition of similarity,

△APC∼△DPB

(ii) △APC∼△DPB

So, sides are proportional

⇒

DP

AP

=

PB

CP

∴AP×PB=CP×DP