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?


fig is not given in this question.


plz answer this question it is very important for me

Answer 1

Given : l∥m

Transversal p intersects l & m at A & C respectively. Bisector of ∠ PAC & ∠ QCA meet at B. And, bisector of ∠ SAC & ∠ RCA meet at D.

To prove : ABCD is a rectangle.

Proof :

We know that a rectangle is a parallelogram with one angle 90°.

For l∥m and transversal p

∠PAC=∠ACR

So, 1/2

∠PAC=1/2

∠ACR

So, ∠BAC=∠ACD

For lines AB and DC with AC as transversal ∠BAC & ∠ACD are alternate angles, and they are equal.

So, AB∥DC.

Similarly, for lines BC & AD, with AC as transversal ∠BAC & ∠ACD are alternate angles, and they are equal.

So, BC∥AD.

Now, In ABCD,

AB∥DC & BC∥AD

As both pair of opposite sides are parallel, ABCD is a parallelogram.

Also, for line l,

∠PAC+∠CAS=180° 1/2

∠PAC+ 1/2

∠CAS=90°

∠BAC+∠CAD=90°

∠BAD=90° .

So, ABCD is a parallelogram in which one angle is 90°.

Hence, ABCD is a rectangle.