Question

Prove that the quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

Answer 1

Hint: In this given question, we can use the fact that adjacent angles of a parallelogram are supplementary meaning their sum is equal to ${{180}^{\circ }}$. Then we can use the concept of Vertically Opposite Angles (VOA) as equal to prove that each angle of the quadrilateral formed is a right angle, hence making it a rectangle.

Complete step-by-step answer:
In this given question, we are asked to prove that the quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

Here, we are going to the fact that adjacent angles of a parallelogram are supplementary meaning their sum is equal to ${{180}^{\circ }}$.
Also, we are going to use the angle sum property of triangles which gives us that the sum of all the angles of a triangle is equal to ${{180}^{\circ }}$.
The process of solving is as follows:
In parallelogram ABCD, as adjacent sides are supplementary so,
$\begin{align}
  & \angle B+\angle C={{180}^{\circ }} \\
 & \Rightarrow \dfrac{1}{2}\left( \angle B+\angle C \right)=\dfrac{1}{2}\times {{180}^{\circ }} \\
 & \Rightarrow \dfrac{1}{2}\angle B+\dfrac{1}{2}\angle C={{90}^{\circ }}.............(1.1) \\
\end{align}$
As, angle bisectors bisect the angles into two equal halves,
$\angle QBC=\dfrac{1}{2}\angle B\text{ and }\angle \text{QCB=}\dfrac{1}{2}\angle C...........(1.2)$
Now, in $\Delta BQC$,
$\angle QBC+\angle QCB+\angle BQC={{180}^{\circ }}$ (by angle sum property of triangles)
$\Rightarrow \dfrac{1}{2}\angle B+\dfrac{1}{2}\angle C+\angle BQC={{180}^{\circ }}$ (from 1.2)
$\Rightarrow {{90}^{\circ }}+\angle BQC={{180}^{\circ }}$(From 1.1)
$\Rightarrow \angle BQC={{90}^{\circ }}.............(1.3)$
Now, as vertically opposite angles are equal,
$\angle BQC=\angle PQR............(1.4)$
From 1.3 and 1.4, we get,
$\angle PQR={{90}^{\circ }}$
Similarly, we can also obtain that,
$\angle QRS=\angle PSR=\angle SRQ={{90}^{\circ }}$
So, we get,
$\angle PQR=\angle QRS=\angle PSR=\angle SRQ={{90}^{\circ }}$
As all the four angles of the quadrilateral are right angles, we can conclude that it is a rectangle.
Therefore, we have proved that the quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

Note: In this sort of question, we may have also used another triangle in order to get the basis as proof as an example instead of $\Delta BQC$. Then we may have followed the same procedure and would have arrived at the same conclusion.