Question

If the angle bisectors of a parallelogram form another quadrilateral, then this quadrilateral will be a

• A. Square
• B. Rhombus
• C. rectangle
• D. Parallelogram

Answer 1

The correct option is C

rectangle

If the angle bisector of a parallelogram form another quadrilateral, then this quadrilateral will be a rectangle
Given:- $ABCD$ is a parellelogram, $AP$, $BR$, $CR$ and $DP$ are angle bisectors.
To prove:- $PQRS$ is a rectangle.
Proof:- In parellelogram $ABCD$, AD$\parallel$CB
Therefore $\angle$ $A$ + $\angle$ $B$=${180}^{\circ }$. [ Sum of Angles on the same side of transversal is ${180}^{\circ }$]
$\Rightarrow$ $\frac{1}{2}$($\angle$ $A$ + $\angle$ $B$)= $\frac{\left({180}^{\circ }\right)}{2}$
$\Rightarrow$ $\frac{1}{2}$($\angle$ $A$ + $\angle$ $B$)=${90}^{\circ }$ .....(1)
By Angle Sum Property in $△$ $ASB$,
$\angle$ ABS + $\angle$ BSA + $\angle$ SAB= ${180}^{\circ }$
$\Rightarrow$ $\frac{1}{2}$ $\angle$ B + $\angle$ BSA + $\frac{1}{2}$ $\angle$ A= ${180}^{\circ }$( given that AP and BR are angle bisectors)
$\Rightarrow$${90}^{\circ }$+ $\angle$ BSA =${180}^{\circ }$ (From (1))
$\Rightarrow$$\angle$ BSA = ${90}^{\circ }$
Similarly , $\angle$ BRC= $\angle$CQD= $\angle$APD =${90}^{\circ }$
$\therefore$ A quadrilateral $PQRS$ in which all angles are right angles is a rectangle.