Question

Prove that the bisector of interior angles of a parallelogram form a rectangle.

Answer 1

STEP 1 : Assumptions

Assume that $ABCD$ is a parallelogram.

Let $P,Q,R,S$ be the point of intersection of bisectors of $\angle A$ and $\angle B$, $\angle B$ and $\angle C$, $\angle C$ and $\angle D$, $\angle D$ and $\angle A$, respectively as shown in figure.

STEP 2 : Finding the value of $\mathbf{\angle }\mathit{D}\mathit{S}\mathit{A}$

From the triangle, $ASD$, we can observe that $DS$ bisects $\angle D$ and $AS$ bisects $\angle A$.

Therefore,

$\angle ADS+\angle DAS=\frac{1}{2}\angle D+\frac{1}{2}\angle A$

$\angle ADS+\angle DAS=\frac{1}{2}\left(\angle A+\angle D\right)$

$\angle ADS+\angle DAS=\frac{1}{2}\left(180°\right)$ [Since $\angle A$ and $\angle D$ are the interior angles on the same side of the transversal]

Hence,

$\angle ADS+\angle DAS=90°$

Using the angle sum property of a triangle, we can write:

$\angle DAS+\angle ADS+\angle DSA=180°$

Now, substitute $\angle ADS+\angle DAS=90°$ in the above equation, we get

$90°+\angle DSA=180°$

$\angle DSA=180°-90°$

$\angle DSA=90°$

STEP 3 : Finding the value of $\mathbf{\angle }\mathit{A}\mathit{P}\mathit{B}\mathbf{,}\mathbf{}\mathbf{\angle }\mathit{D}\mathit{R}\mathit{C}\mathbf{,}\mathbf{}\mathbf{\angle }\mathit{P}\mathit{S}\mathit{R}$ and $\mathbf{\angle }\mathit{P}\mathit{Q}\mathit{R}$

Since, $\angle PSR$ is being vertically opposite to $\angle DSA$,

We can say $\angle PSR=90°$

Likewise, it can be shown that $\angle APB=90°$ or $\angle APB=90°$

Similarly, $\angle PQR=90°$ and $\angle SRQ=90°$.

We have proved that $\angle PSR=\angle PQR=90°$ and $\angle APB=\angle CRD=90°$.

As, both the pairs of opposite angles are equal to $90°$, we can conclude that $PQRS$ is a rectangle.

Hence, proved.