Question

Show that the bisectors of angles of a parallelogram form a rectangle

Answer 1

Given: A parallelogram in which bisector of angle $A,B,C,D$ intersect at $P,Q,R,S$ to form a quadrilateral $PQRS$.
To prove: Quadrilateral PQRS is a rectangle.
Proof: Since $ABCD$ is a parallelogram.
Therefore, $AB\parallel DC$.

Now, $AB\parallel DC$, and transversal $AD$ cuts them, so we have
$\begin{array}{cc}& \angle A+\angle D={180}^{\circ }\\ & \frac{1}{2}\angle A+\frac{1}{2}\angle D=\frac{{180}^{\circ }}{2}\\ & \angle DAS+\angle ADS={90}^{\circ }\end{array}$
But in $△ASD$, we have
$\begin{array}{cc}& \angle ADS+\angle DAS+\angle ASD={180}^{\circ }\\ & {90}^{\circ }+\angle ASD={180}^{\circ }\\ & \angle ASD={90}^{\circ }\end{array}$
$\angle RSP=\angle ASD$...(vertically opposite angle)
$\angle RSP={90}^{\circ }$
Similarly, we can prove that
$\angle SRQ={90}^{\circ },\angle RQP={90}^{\circ }\text{and}\angle QPS={90}^{\circ }$
Thus, PQRS is a quadrilateral each of whose angle is ${90}^{\circ }$.
Hence, PQRS is a rectangle.