Question

please help me out
please​

Answer 1

Given :

• QT and RT is bisector of ∠PQR and ∠PRS.

To Prove :

• ∠QTR = 1/2 ∠QPR

Solution :

$\longmapsto\tt{Let\:\angle{PQR}=x}$

$\longmapsto\tt{Let\:\angle{PRS}=y}$

Now , In Δ PQR :

• ∠PRS is exterior angle

$\longmapsto\tt{\angle{PRS}=\angle{QPR}+\angle{PQR}}$

$\longmapsto\tt{y=\angle{P}+x}$

$\longmapsto\tt\bold{y-x=\angle{P}}$

Now ,In Δ TQR :

• ∠TRS is exterior angle

$\longmapsto\tt{\angle{TRS}=\angle{TQR}+\angle{QTR}}$

$\longmapsto\tt{\dfrac{y}{2}=\dfrac{x}{2}+\angle{T}}$

$\longmapsto\tt{\dfrac{y}{2}-\dfrac{x}{2}=\angle{T}}$

$\longmapsto\tt{\dfrac{\angle{P}}{2}=\angle{T}}$

$\longmapsto\tt{\angle{T}=\dfrac{1}{2}\angle{P}}$

$\longmapsto\tt\bold{\angle{QTR}=\dfrac{1}{2}\angle{QPR}}$

HENCE PROVED