Question

if two parallel line are intersected by a transversal prove that the bisector of the two pair of interior angle enclosed a rectangles ​

Answer 1

Answer:

Answer

Given: Two parallel lines AB and CD and a transversal EF intersect them at G and H respectively. GM, HM, GL and HL are the bisectors of the two pairs of interior angles.

To Prove: GMHL is a rectangle.

Proof:

∵AB∥CD

∴∠AGH=∠DHG (Alternate interior angles)

⇒

2

1

∠AGH=

2

1

∠DHG

⇒∠1=∠2

(GM & HL are bisectors of ∠AGH and ∠DHG respectively)

⇒GM∥HL

(∠1 and ∠2 from a pair of alternate interior angles and are equal)

Similarly, GL∥MH

So, GMHL is a parallelogram.

∵AB∥CD

∴∠BGH+∠DHG=180

o

(Sum of interior angles on the same side of the transversal =180

o

2 1

∠BGH+

2

1

∠DHG=90

⇒∠3+∠2=90 _ ( 3)

(GL & HL are bisectors of ∠BGH and ∠DHG respectively).

In ΔGLH,∠2+∠3+∠L=180

⇒90

+∠L=180

Using (3)

⇒∠L=180

−90

⇒∠L=90

Thus, in parallelogram GMHL, ∠L=90

Hence, GMHL is a rectangle.

Answer 2

Step-by-step explanation:

AB and CD are two parallel lines .

EF is a transversal line which intersects the two parallel lines at points G and H and the bisector of two pares of interior angle intersect in L and M.

And this forms a parlellogram=rectangle