# in a quadrilateral pqrs ,the bisectors of angle R and angle S meet at point T.Prove that angle P+ angle Q= 2angle RTS

We have a quadrilateral PQRS , where bisectors of angle R and angle S meet at point T. $\angle$ P +
$\angle$Q + $\angle$R + $\angle$S = 360$°$
$\angle$ P + $\angle$Q =  360$°$ - ($\angle$R + $\angle$S)          --------------- (1)

In $∆$RTS ,
$\angle$RTS + $\angle$TSR  + $\angle$SRT = 180$°$             ---------------  (2)
Given,
 $\angle$TSR    = $\frac{\angle S}{2}$
$\angle$SRT  = $\frac{\angle R}{2}$
After substitute this in  equation (2) , we get

$\angle$RTS + $\frac{\angle S}{2}$ +$\frac{\angle R}{2}$ =  180$°$
⇒ 2$\angle$RTS  +$\angle$S  +$\angle$R   = 360$°$
2$\angle$RTS = 360$°$ - ($\angle$R + $\angle$S)
From equation number (1)
2$\angle$RTS =   $\angle$ P + $\angle$Q                                         (Hence proved)

We have a quadrilateral PQRS , where bisectors of angle R and angle S meet at point T. $\angle$ P +
$\angle$Q + $\angle$R + $\angle$S = 360$°$
$\angle$P + $\angle$Q = 180$°$                                  -------------(1)
$\angle$R + $\angle$S =   180$°$                                                       ( Angles on one common line makes 180$°$)
In $∆$RTS ,
$\angle$RTS + $\angle$TSR  + $\angle$SRT = 180$°$             ---------------  (2)
$\angle$TSR  + $\angle$SRT ​  = 90$°$    (These angle form by bisectors on $\angle$R and $\angle$S and we know $\angle$R + $\angle$S =   180$°$ ​)

Substitute this in  equation (2) , we get
$\angle$RTS + 90$°$ = 180$°$
$\angle$RTS    = 90$°$
Multiply by 2 on both side , we get
2$\angle$RTS = 180$°$
After substitute in equation (1) we get

$\angle$P + $\angle$Q = 2$\angle$RTS​                  (Hence proved)

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