
In a fig. the side QR of is produced to a point S. If the bisectors of $\angle PQR$ and $\angle PRS$ meet at point T, then prove that $\angle QRT = \dfrac{1}{2}\angle QPR$.
Answer
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Hint: In this question we will use the concept of the angle sum property of a triangle which states that if a side of a triangle is produced then the exterior angle so formed is equal to the some of the two interior opposite angles.
Complete step by step solution:
We know that the exterior angle of a triangle is greater than either of its interior opposite angles.
Given that, the bisectors of $\angle PQR$ and $\angle PRS$ meets at point T.
To prove: $\angle QRT = \dfrac{1}{2}\angle QPR$.
Proof:
TQ is the bisector of $\angle PQR$.
So, $\angle PQT = \angle TQR = \dfrac{1}{2}\angle PQR$ ………(i)
Also,
TR is the bisector of $\angle PRS$.
So, $\angle PRT = \angle TRS = \dfrac{1}{2}\angle PRS$. ………..(ii)
Now, in $\vartriangle PQR$
$\angle PRS$ is the external angle and we know that the external angle is the sum of two interior opposite angles.
Therefore,
$\angle PRS = \angle QPR + \angle PQR$ …………(iii)
Similarly, in $\vartriangle TQR$
$\angle TRS$ is the external angle and we know that the external angle is the sum of two interior opposite angles.
$\angle TRS = \angle TQR + \angle QTR$ …………..(iv)
Now we will put the values from equation (i) and (ii) in equation (iv), and we get
$\dfrac{1}{2}\angle PRS = \dfrac{1}{2}\angle PQR + \angle QTR$
And by putting the value of $\angle PRS$ from equation (iii), we get
$\dfrac{1}{2}\left( {\angle QPR + \angle PQR} \right) = \dfrac{1}{2}\angle PQR + \angle QTR$.
By solving this step by step,
$
\dfrac{1}{2}\angle QPR + \dfrac{1}{2}\angle PQR = \dfrac{1}{2}\angle PQR + \angle QTR \\
\dfrac{1}{2}\angle QPR + \dfrac{1}{2}\angle PQR - \dfrac{1}{2}\angle PQR = \angle QTR \\
\dfrac{1}{2}\angle QPR = \angle QTR \\
$
We get, $\angle QTR = \dfrac{1}{2}\angle QPR.$
Hence proved $\angle QRT = \dfrac{1}{2}\angle QPR$.
Note: whenever we ask such types of questions, we have to remember some basic points of angle sum property of a triangle. First we have to know what is given and what we have to prove. Then according to the given statements, we will make some statements and equations and by putting the appropriate values we will solve them. After solving, we will get the required answer.
Complete step by step solution:
We know that the exterior angle of a triangle is greater than either of its interior opposite angles.
Given that, the bisectors of $\angle PQR$ and $\angle PRS$ meets at point T.
To prove: $\angle QRT = \dfrac{1}{2}\angle QPR$.
Proof:
TQ is the bisector of $\angle PQR$.
So, $\angle PQT = \angle TQR = \dfrac{1}{2}\angle PQR$ ………(i)
Also,
TR is the bisector of $\angle PRS$.
So, $\angle PRT = \angle TRS = \dfrac{1}{2}\angle PRS$. ………..(ii)
Now, in $\vartriangle PQR$
$\angle PRS$ is the external angle and we know that the external angle is the sum of two interior opposite angles.
Therefore,
$\angle PRS = \angle QPR + \angle PQR$ …………(iii)
Similarly, in $\vartriangle TQR$
$\angle TRS$ is the external angle and we know that the external angle is the sum of two interior opposite angles.
$\angle TRS = \angle TQR + \angle QTR$ …………..(iv)
Now we will put the values from equation (i) and (ii) in equation (iv), and we get
$\dfrac{1}{2}\angle PRS = \dfrac{1}{2}\angle PQR + \angle QTR$
And by putting the value of $\angle PRS$ from equation (iii), we get
$\dfrac{1}{2}\left( {\angle QPR + \angle PQR} \right) = \dfrac{1}{2}\angle PQR + \angle QTR$.
By solving this step by step,
$
\dfrac{1}{2}\angle QPR + \dfrac{1}{2}\angle PQR = \dfrac{1}{2}\angle PQR + \angle QTR \\
\dfrac{1}{2}\angle QPR + \dfrac{1}{2}\angle PQR - \dfrac{1}{2}\angle PQR = \angle QTR \\
\dfrac{1}{2}\angle QPR = \angle QTR \\
$
We get, $\angle QTR = \dfrac{1}{2}\angle QPR.$
Hence proved $\angle QRT = \dfrac{1}{2}\angle QPR$.
Note: whenever we ask such types of questions, we have to remember some basic points of angle sum property of a triangle. First we have to know what is given and what we have to prove. Then according to the given statements, we will make some statements and equations and by putting the appropriate values we will solve them. After solving, we will get the required answer.
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