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Question-164331




Question Number 164331 by HongKing last updated on 16/Jan/22
Commented by mr W last updated on 16/Jan/22
see Q163656  ((BC)/(AD))=((PB^2 −PC^2 )/(PA^2 −PD^2 ))=1   ⇒AD=BC ⇒rectangle
$${see}\:{Q}\mathrm{163656} \\ $$$$\frac{{BC}}{{AD}}=\frac{{PB}^{\mathrm{2}} −{PC}^{\mathrm{2}} }{{PA}^{\mathrm{2}} −{PD}^{\mathrm{2}} }=\mathrm{1}\: \\ $$$$\Rightarrow{AD}={BC}\:\Rightarrow{rectangle} \\ $$
Commented by Tawa11 last updated on 16/Jan/22
Weldone sir
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

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