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 ABCD.
In $\triangle ASD$,
Since DS bisects $\angle D$ and AS bisects $\angle A$,
therefore,
$\angle DAS$ + $\angle ADS$ = $\frac 12$$\angle A$ + $\frac 12$$\angle D$
= $\frac 12$ ($\angle A$ + $\angle D$)
= $\frac 12 \times 180^o$ ($\angle A$ and $\angle D$ are interior angles on the same side of the transversal)
= 90o
Also, $\angle DAS$ + $\angle ADS$ + $\angle DSA$ = 180o (Angle sum property of a triangle)
or, 90o + $\angle DSA$ = 180o
or, $\angle DSA$ = 90o
So, $\angle PSR$ = 90o (Being vertically opposite to $\angle DSA$)
Similarly, it can be shown that $\angle APB$ = 90o or $\angle SPQ$ = 90o (as it was shown for $\angle DSA$).
Similarly, $\angle PQR$ = 90o and $\angle SRQ$ = 90o.
So, PQRS is a quadrilateral in which all angles are right angles.
We have shown that $\angle PSR$ = $\angle PQR$ = 90o and $\angle SPQ$ = $\angle SRQ$ = 90o.
So both pairs of opposite angles are equal.
Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90o and so, PQRS is a rectangle.
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