1.

Please answer.will mark you brainliest​

Answer»

GIVEN:-

  • The SIDE of QR of ∆PQR is Produced to a POINT S.

  • if the Bisector of \rm{\angle{PQR}\:and\:\angle{PRS}} meet at the Point T.

TO PROVE

  • \rm{\angle{QTR}=\dfrac{1}{2}\angle{QPR}}.

PROPERTY USED:-

  • Exterior angle Property i.e the Sum of TWO Opposite INTERIOR angle is equal to the external angle.

Now,

In ∆QTR

\implies\rm{\angle{TQR}+\angle{QTR}=\angle{TRS}}(Exterior angle Property)

\implies\rm{\angle{QTR}=\angle{TRS}-\angle{TQR}}.......1

Now,

In ∆PQR

\implies\rm{\angle{PQR}+\angle{QPR}=\angle{PRS}}(Exterior angle Property).

[TR is the Bisector of

  • \rm{\dfrac{1}{2}\angle{PQR}=\angle{TQR}=\angle{PQR}=\angle{2TQR}}.......2

  • Similarly, \rm{\angle{PRS}=\angle{2TRS}}..........3

So, From 2 and 3

\implies\rm{\angle{2TQR}+\angle{QPR}=\angle{2TRS}}

\implies\rm{\angle{QPR}=2(\angle{TRS}-\angle{TQR})}

Substituting the value of eq1

\implies\rm{\angle{QPR}=2(\angle{QTR})}

or,

\implies\rm{\dfrac{1}{2}\angle{QPR}=\angle{QTR}}


Discussion

No Comment Found

Related InterviewSolutions