(P & S are 2 pts on line l and Q & R are d 2 pts on line m)
It is given that PS//QR(l//m) and transversal t intersects them at point A and C respectively.
The bisectors of angle PAC and angle ACQ intersect at B and bisectors of angle ACR and SAC intersect at D.
To show that: Quadrilateral ABCD is a rectangle
Now, angle PAC= angle ACR(alternate angles as l//m and t is a transversal)
i.e angle BAC= angle ACD
These form a pair of alternate angles for lines AB and DC with AC as transversal and they are also equal
Similarly BC//AD (Considering angle ACB and angle CAD)
Therefore quadrilateral ABCD is a //gm
Also angle PAC + angle CAS= 180 degree(linear pair)
1/2 angle PAC + 1/2 angle CAS= 1/2 x 180 degree
or angle BAC + angle CAD = 90 degree
or angle BAD= 90 degree
So ABCD is a //gm in which 1 angle is 90 degree
Therefore ABCD is a rectangle.