In Fig., two chords AB and CD intersect each other at the point P. Prove that :
(i) $△ A P C \sim △ D P B$
(ii) $A P . P B = C P . D P$

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Solution

(i) Given : In $△ A P C$ and $△ D P B$ ,
$∠ A P C = ∠ D P B$ ...[Vert. opp. ∠s]
$∠ C A P = ∠ B D P$ ...[Angles subtended by the same arc of a circle are equal]
$∴$ By AA-condition of similarity,
$△ A P C \sim △ D P B$

(ii) $△ A P C \sim △ D P B$
So, sides are proportional
$\Rightarrow \frac{A P}{D P} = \frac{C P}{P B}$

$∴ A P \times P B = C P \times D P$

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In Fig., two chords AB and CD intersect each other at the point P. Prove that :(i) △APC∼△DPB(ii) AP.PB=CP.DP

(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 ⇒APDP=CPPB ∴AP×PB=CP×DP

In Fig., two chords AB and CD intersect each other at the point P. Prove that :
(i) $△ A P C \sim △ D P B$
(ii) $A P . P B = C P . D P$