Question
Two parallel lines $l$ and $m$ are intersected by a transversal $p$. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
✓
Answer
Here, it is given that $QR \parallel PS$ and transversal $p$ intersects them at points $A$ and $C$ respectively.
The bisectors of $\angle ACR$ and $\angle SAC$ intersect at $D$ and the bisectors of $\angle PAC$ and $\angle ACQ$ intersect at $B.$
We have to prove that quadrilateral$\text{ABCD}$ is a rectangle.
From the given figure, we have
$\angle PAC = \angle ACR [$Alternate angles as $l \parallel m$ and $p$ is a transversal$]$
So, $\frac{1}{2} \angle PAC = \frac{1}{2} \angle ACR$
i.e., $\angle BAC = \angle ACD$
These form a pair of alternate angles for lines $AB$ and $DC$ with $AC$ as a transversal and they are equal also.
So$, AB \parallel DC$
Similarly $, BC \parallel AD [$Considering $\angle ACB$ and $\angle CAD]$
Therefore, quadrilateral $\text{ABCD}$ is a parallelogram
Also, $\angle PAC + \angle CAS = 180^\circ [$ by the property of Linear pair$]$
So, $\frac{1}{2} \angle PAC + \frac{1}{2} \angle CAS = \frac{1}{2} \times 180^\circ = 90^\circ$
or, $\angle BAC + \angle CAD = 90^\circ$
or, $\angle BAD = 90^\circ$
So, $\text{ABCD}$ is parallelogram in which one angle $(\angle BAD )$ is $90^\circ$.
Therefore, $\text{ABCD}$ is a rectangle.
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