Question
Show that the bisectors of angles of a parallelogram form a rectangle.
✓
Answer
Let $P , Q , R$ and S be the points of intersection of the bisectors of $\angle A$ and $\angle B, \angle B$ and $\angle C, \angle C$ and $\angle D$, and $\angle D$ and $\angle A$ respectively of parallelogram $A B C D$.
$\text { In } \triangle A S D$
Since DS bisects $\angle D$ and AS bisects $\angle A$, therefore,
$\angle D A S+\angle A D S=\frac{1}{2} \angle A+\frac{1}{2} \angle D$
$=\frac{1}{2}(\angle A+\angle D)$
$=\frac{1}{2} \times 180^{\circ} \text { ( } \angle A \text { and } \angle D$ are interior angles on the same side of the transversal)
$=90^{\circ}$
Also, $\angle D A S+\angle A D S+\angle D S A=180^{\circ}$ (Angle sum property of a triangle)
$\text { or, } 90^{\circ}+\angle D S A=180^{\circ}$
or, $\angle D S A=90^{\circ}$
So, $\angle P S R=90^{\circ}$ (Being vertically opposite to $\angle D S A$ )
Similarly, it can be shown that $\angle A P B=90^{\circ}$ or $\angle S P Q=90^{\circ}$ (as it was shown for $\angle D S A$ ).
Similarly, $\angle P Q R=90^{\circ}$ and $\angle S R Q=90^{\circ}$.
So, PQRS is a quadrilateral in which all angles are right angles.
We have shown that $\angle P S R=\angle P Q R=90^{\circ}$ and $\angle S P Q=\angle S R Q=90^{\circ}$.
So both pairs of opposite angles are equal.
Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is $90^{\circ}$ and so, PQRS is a rectangle.
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