Question
In Fig. it is given that $\text{RT}=\text{TS},\angle1=2\angle2$ and $\angle4=2\angle3.$ prove that $\triangle\text{RBT}\cong\triangle\text{SAT}.$
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Answer
Here, $\angle1=2\angle2$ and $\angle4=2\angle3$ $\big[\therefore$ Exterior angle = sum of opposite interior angles $\big]$
$\angle1=\angle4$ [vertically opposite angle]
$\therefore2\angle2=2\angle3$
$\Rightarrow\angle2=\angle3$
Now $\text{RT}=\text{TS}$ [given]
$\Rightarrow\angle\text{TRS}=\angle\text{TSR}$ [Angle opposite to equal sides are equal]
$\therefore\angle\text{TRS}-\angle2=\angle\text{TSR}-\angle3$
$\Rightarrow\angle\text{TRB}=\angle\text{TSA}$
Now in $\triangle\text{RBT}$ and $\triangle\text{SAT}$
$\angle\text{T}=\angle\text{T}$ [common]
$\angle\text{TRB}=\angle\text{TSA}$ [proved earlier]
$\text{RT}=\text{TS}$ [given]
By ASA conguence criterion $\triangle\text{RBT}\cong\triangle\text{SAT}$
$\angle1=\angle4$ [vertically opposite angle]
$\therefore2\angle2=2\angle3$
$\Rightarrow\angle2=\angle3$
Now $\text{RT}=\text{TS}$ [given]
$\Rightarrow\angle\text{TRS}=\angle\text{TSR}$ [Angle opposite to equal sides are equal]
$\therefore\angle\text{TRS}-\angle2=\angle\text{TSR}-\angle3$
$\Rightarrow\angle\text{TRB}=\angle\text{TSA}$
Now in $\triangle\text{RBT}$ and $\triangle\text{SAT}$
$\angle\text{T}=\angle\text{T}$ [common]
$\angle\text{TRB}=\angle\text{TSA}$ [proved earlier]
$\text{RT}=\text{TS}$ [given]
By ASA conguence criterion $\triangle\text{RBT}\cong\triangle\text{SAT}$
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