In the given figure, QR is parallel to AB and DR is parallel to QB.
Prove that : $P Q^{2} = P D \times P A$ .

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Solution

Given, QR is parallel to AB. Using Basic proportionality theorem,

$fraction numerator P Q over denominator P A end fraction equals fraction numerator P R over denominator P B end fraction minus negative negative negative open parentheses 1 close parentheses$

Also, DR is parallel to QB. Using Basic proportionality theorem,

$fraction numerator P D over denominator P Q end fraction equals fraction numerator P R over denominator P B end fraction minus negative negative open parentheses 2 close parentheses$

From (1) and (2), we get,

$fraction numerator P Q over denominator P A end fraction equals fraction numerator P D over denominator P Q end fraction P Q squared equals P D cross times P A$

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In the given figure, QR is parallel to AB and DR is parallel to QB. If PD = 4 cm and PA = 9 cm, then PQ = ____.

In the given figure, QR is parallel to AB and DR is parallel to QB. If PD = 4 cm and PA = 9 cm, then find PQ.

P A ⊥ A B, Q B ⊥ A B

P A = Q B

△ O A P ≅ O B Q

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