Quadrilaterals — Maths STD 9 — Question

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 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 ABCD is a parallelogram
Also, $\angle$PAC + $\angle$CAS = 180^o   [ by the property of Linear pair ]
So, $\frac{1}{2}$$\angle$PAC + $\frac{1}{2}$$\angle$CAS = $\frac{1}{2}$ $\times$ 180^o = 90^o
or, $\angle$BAC + $\angle$CAD = 90^o
or, $\angle$BAD = 90^o
So, ABCD is parallelogram in which one angle ( $\angle$BAD ) is 90^o.
Therefore, ABCD is a rectangle.