In this figure Prove $\displaystyle "Angle" RQS = 2 \times "Angle" BDC$
"Angle" QRD = "Angle" SRB,
"Angle" RSD = "Angle" ASC and
DRB, DSC, QSA are straight lines.
Let $\displaystyle \widehat{QRS}=\widehat{BRS}=x$ and $\displaystyle \widehat{RSD}=\widehat{ASC}=y$
$\displaystyle \widehat{QRS}=180-2x$
$\displaystyle \widehat{RQS}=180-\widehat{QRS}-\widehat{QSR}=$
$\displaystyle =180-2y-(180-2x)=2(x-y)$
$\displaystyle \widehat{BDC}=\widehat{RDS}=180-\widehat{DSR}-\widehat{SRD}=$
$\displaystyle =180-x-y-(180-2x)=x-y$
So, $\displaystyle \widehat{QRS}=2\widehat{BDC}$