Question

In the given fig, \[PR > PQ\] and PS bisects$\angle QPR$. Prove that$\angle PSR > \angle PSQ$.

Answer 1

Hint: To solve this question use the theorem of exterior angle of a triangle which states that exterior angle is equal to the sum of other two interior angles of triangle which is opposite to the adjacent angle of it.

Complete Step-by-step Solution
Given: In $\Delta PQR$, PS is a bisector of $\angle QPR$ and side PR is larger than side PQ.
PS is a bisector of $\angle QPR$ so it divides the angle in two equal parts.
$\angle QPS = \angle RPS = x$
From$\Delta PQS$, $\angle PSR$ is the exterior angle of the triangle and we know that the exterior angle of a triangle is equal to the sum of interior opposite angles.
\[\begin{array}{c}
\angle PSR = \angle QPS + \angle PQS\\
\angle PSR = x + \angle PQS
\end{array}\]…….(1)
Similarly from$\Delta PSR$, $\angle PSQ$ is the exterior angle of the triangle and we know that the exterior angle of a triangle is equal to the sum of interior opposite angles.
\[\begin{array}{l}
\angle PSQ = \angle RPS + \angle PRS\\
\angle PSQ = x + \angle PRS
\end{array}\]………(2)
So the angle opposite to these sides follows the same relation as,
$\angle PQR > \angle PRQ$
These angles can be written as,
$\angle PQS > \angle PRS$
Add angle x on both sides of the equation.
$\angle PQS + x > \angle PRS + x$……(3)
Comparing equation (1), (2) and (3).
$\angle PSR > \angle PSQ$

Note: The relation of sides is given in question cannot be used directly to prove the question. First we need to solve the relation and after that we can prove the question by their indirect relations with each other. In the solution, the theorem of exterior angle of a triangle is used in $\Delta PQS$ and $\Delta PSR$ to prove $\angle PSR > \angle PSQ$. The median PS divides the triangle $\Delta PQR$ in two smaller triangles and the area of both the triangles are the same.