Question

Two parallel lines l and m are intersected by a transversal 't'. Show that quadrilateral formed by bisectors of interior angles is a rectangle.

Answer 1

(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

So AB//DC

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.

Answer 2

please see diagram.

LL and MM are parallel lines.  tt is a transverse line.  They intersect at A and D.    The two angular bisectors for the acute and obtuse angles are drawn. They intersect at B and C.

We have to prove that ABCD is a rectangle. 

Angle BAD = angle  BAL  (b3 is bisector)
angle DAC = angle CAL  (b1 is bisector)
angles LAB + angle BAD + Angle DAC + angle CAL = 180 deg
    => 2 (angle BAD + angle DAC) = 180 deg
   Hence angle BAC = 90 deg

Similarly,  angle BDC = 90 deg.

angle LAD = angle ADM    (tt  transverse line)
1/2 angle LAD = angle CAD = 1/2 angle ADM = angle ADB
       Hence  b1, b2 are also parallel and  tt is transverse line.
         hence  CD and AB are parallel.
     Hence angles ACD and ABD will be 90 deg also.

Hence ABCD is a rectangle.