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