Question

In figure, the side QR of \[\Delta PQR\] is produced to points. If the bisector of \[\angle PQR\,\,and\,\,\angle PRS\]meet at point T, then prove that
\[\angle QTR\,\,=\,\,\dfrac{1}{2}\angle QPR\]

Answer 1

Hint:In this problem use the concept of the exterior angle of the triangle equals to the sum of two interior angles.
Use the angle sum property of the triangle.


Complete step by step solution:
Given that
(i) Side QR of \[\Delta PQR\]is produced to a point S.
(ii) Bisector of \[\angle PQR\]and \[\angle PRS\]meet at point T
\[\angle TRS\,\,=\,\,\angle TQR\,\,+\,\angle QTR\]
(Exterior angle of a triangle equals to the sum of the two interior opposite angles)
\[\angle QTR\,\,=\,\,\angle TRS\angle TQR\,\,\,\,........(1)\]
(According to the given figure)
\[\angle SRP\,\,=\,\,\angle QPR\,\,+\,\,\angle QPR\]
\[\Rightarrow \,\,2\angle TRS\,\,\,=\,\,\angle QPR\,\,+\,\,2\angle TQR\]
TR is a bisector of \[\angle SRP\]and QT is angle bisector of \[\angle PQR\]
\[\Rightarrow \,\angle QPR\,\,=\,\,2\angle TRS2\,\,\angle TQR\]
Taken \[2\]is the common
Then
\[\angle QPR\,\,=\,\,(\angle TRS\angle TQR)\]
\[\angle TRS\angle TQR\,=\,\dfrac{1}{2}\angle QPR\]
Substitute the value of the \[\angle QTR\] from the equation\[(1)\]
\[\angle TRS\angle TQR=\angle QTR\]
So that
\[\angle QTR\,\,=\,\,\dfrac{1}{2}\angle QPR\]
Hence proved
Note:There exists an alternate method to solve this,this problem can also be solved by the exterior angle bisector method.