Question

In the given figure, $\angle BAC=\angle BDC={90}^{\circ }$. If AC and DB intersect at P, then:

• A. $AP\times PC=BP\times PD$
• B. $AP\times BP=PC\times PD$
• C. $AP\times PD=PC\times BP$
• D. $Noneofthese$

Answer 1

The correct option is A $AP\times PC=BP\times PD$

$\Delta BAP$ ~ $\Delta CDP(\because \angle A=\angle Dand\angle APB=\angle DPC)$
$\therefore \frac{AB}{CD}=\frac{AP}{DP}=\frac{BP}{CP}$
$\therefore AP\times PC=BP\times DP$