Question

two Parallel Lines l and M are intersected by a transversal p show that quadrilateral formed by the bisector of interior angle is a rectangle

Answer 1

but angle LGH+ ANGLE LHG+ Angle GLH=180°
(ANGLE SUM PROPERTY)
90°+angle GLH=180°
(°.° angle LHG + angle LGH =90° )
Angle GLH = 90°
THUS in a parallelogram we have angle GLH =90°
hence , GMHL is a rectangle

Answer 2

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 90

o

.

For l∥m and transversal p

∠PAC=∠ACR

So,

2

1

∠PAC=

2

1

∠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

o

2

1

∠PAC+

2

1

∠CAS=90

o

∠BAC+∠CAD=90

o

∠BAD=90

o

.

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

o

.

Hence, ABCD is a rectangle.

solution