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,
$\begin{array}{rl}& \mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }\\ & \Rightarrow {\displaystyle \frac{1}{2}}(\mathrm{\angle }B+\mathrm{\angle }C)={\displaystyle \frac{1}{2}}\times {180}^{\circ }\\ & \Rightarrow {\displaystyle \frac{1}{2}}\mathrm{\angle }B+{\displaystyle \frac{1}{2}}\mathrm{\angle }C={90}^{\circ }.............(1.1)\end{array}$
As, angle bisectors bisect the angles into two equal halves,
$\mathrm{\angle }QBC={\displaystyle \frac{1}{2}}\mathrm{\angle }B\text{ and }\mathrm{\angle }\text{QCB=}{\displaystyle \frac{1}{2}}\mathrm{\angle }C...........(1.2)$
Now, in $\mathrm{\Delta }BQC$,
$\mathrm{\angle }QBC+\mathrm{\angle }QCB+\mathrm{\angle }BQC={180}^{\circ }$ (by angle sum property of triangles)
$\Rightarrow {\displaystyle \frac{1}{2}}\mathrm{\angle }B+{\displaystyle \frac{1}{2}}\mathrm{\angle }C+\mathrm{\angle }BQC={180}^{\circ }$ (from 1.2)
$\Rightarrow {90}^{\circ }+\mathrm{\angle }BQC={180}^{\circ }$(From 1.1)
$\Rightarrow \mathrm{\angle }BQC={90}^{\circ }.............(1.3)$
Now, as vertically opposite angles are equal,
$\mathrm{\angle }BQC=\mathrm{\angle }PQR............(1.4)$
From 1.3 and 1.4, we get,
$\mathrm{\angle }PQR={90}^{\circ }$
Similarly, we can also obtain that,
$\mathrm{\angle }QRS=\mathrm{\angle }PSR=\mathrm{\angle }SRQ={90}^{\circ }$
So, we get,
$\mathrm{\angle }PQR=\mathrm{\angle }QRS=\mathrm{\angle }PSR=\mathrm{\angle }SRQ={90}^{\circ }$
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 $\mathrm{\Delta }BQC$. Then we may have followed the same procedure and would have arrived at the same conclusion.