Question

If the bisectors of Angles PQR and
angle PRQ of triangle Por meet at a point,
M then prove that 2QMR = 90° +angle p/2​

Answer 1

Given:

$\triangle PQR$ has angle bisectors of $\angle PQR$ and $\angle PRQ$ which intersect at point M.

To prove:

$\angle QMR = 90^\circ + \dfrac{\angle P}2$

Solution:

First of all, please have a look at the attached image for the given $\triangle PQR$.

Angle bisectors of $\angle PQR$ and $\angle PRQ$ intersect at point M.

Now, let us consider $\triangle PQR$.

Sum of all the angles of a triangle is equal to $180^\circ$.

i.e.

$\angle P + \angle PQR + \angle PRQ = 180^\circ\\\Rightarrow \angle PQR + \angle PRQ = 180^\circ - \angle P ...... (1)$

Now, let us consider $\triangle QMR$:

$\angle MQR = \dfrac{1}{2}\angle PQR (\because \text {QM is the angle bisector of } \angle MQR)$

$\angle MRQ = \dfrac{1}{2}\angle PRQ (\because \text {RM is the angle bisector of } \angle MRQ)$

And, Sum of all the angles of a triangle is equal to $180^\circ$.

i.e.

$\angle MQR + \angle MRQ + \angle QMR = 180^\circ\\\Rightarrow \dfrac{1}{2} \angle PQR+\dfrac{1}{2} \angle PRQ + \angle QMR = 180^\circ\\ \Rightarrow \dfrac{1}{2} (\angle PQR+ \angle PRQ) + \angle QMR = 180^\circ\\\\\text{Using Equation (1) :}\\\\\Rightarrow \dfrac{1}{2} (180^\circ - \angle P) + \angle QMR = 180^\circ\\\Rightarrow 90^\circ - \dfrac{1}{2}\angle P + \angle QMR = 180^\circ\\\Rightarrow \angle QMR = 180^\circ -90^\circ + \dfrac{1}{2}\angle P$

$\Rightarrow \angle QMR = 90^\circ + \dfrac{\angle P}{2}$

Hence proved.