Question

A line is parallel to line , q and a transversal l intersects them at A and B,respectively.
Bisectors of interior angles at A and B intersect at C and D. Name the figure ACBD with reasons. ​

Answer 1

Answer:

The figure is a rectangle

Step-by-step explanation:

1- angleA+angleB+angleC+angleD=360

2-angleA+angleB=180 (co-interior)

also angle angle BAC=ACD

since the alternate angles are equal, AC||BD and AD||BC

SO IT IS DEFINITELY A PARALLELOGRAM

angleA=180 (all points that make up angleA)

its half=90

in a parallelogram opposite angles are equal and adjacent angles are =180

so we arrive at the conclusion that ABCD is a parallelogram with all sides=90

therefore ABCD is a rectangle

P.S: it would have been easier had u attached a figure or something. Stay cool!  

Step-by-step explanation:

Answer 2

Two parallel line l and m are intersected by a transversal 'p' show that the quadrilateral formed by bisectors of interior angel is a rectangle. 

ANSWER

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 90o.

For l∥m and transversal p

∠PAC=∠ACR

So, 21∠PAC=21∠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=180o

21∠PAC+21∠CAS=90o

∠BAC+∠CAD=90o

∠BAD=90o.

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

Hence, ABCD is a rectangle.