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In a parallelogram ^@ABCD^@, the bisectors of ^@\angle A^@ and ^@\angle B^@ intersect at ^@S^@, ^@ \angle B ^@ and ^@ \angle C ^@ at ^@R^@, ^@ \angle C^@ and ^@ \angle D ^@ at ^@Q^@ and ^@ \angle D^@ and ^@\angle A^@ at ^@P^@. What kind of a quadrilateral is ^@PQRS^@?
Answer:
^@ Rectangle ^@
- The situation given in the question is represented by the figure below.
- We are given that ^@ABCD^@ is a parallelogram.
@^ \implies DC \parallel AB @^
Also, as the adjacent angles of a parallelogram are supplementary, we have
@^
\begin{aligned}
& \angle A + \angle D = 180^\circ \\
\implies & \dfrac { 1 } { 2 } \angle A + \dfrac { 1 } { 2 } \angle D = 90^\circ \\
\implies & \angle PAD + \angle ADP = 90^\circ \\
\implies & \angle APD = 90^\circ && \text{[Sum of angles of a triangle is } 180^\circ.] \\
\implies & \angle SPQ = 90^\circ && \text{[Vertically opposite angles.]}
\end{aligned}
@^
Similarly, ^@ \angle PQR = 90^\circ, \angle QRS = 90^\circ, ^@ and ^@ \angle PSR = 90^\circ ^@.
Thus, ^@ PQRS ^@ is a quadrilateral each of whose angles is ^@ 90^\circ ^@.
Hence, ^@ PQRS ^@ is a ^@ rectangle ^@.
