Question

Show that quadrilateral formed by the bisectors of interior angles of two parallel lines enclose a rectangle.​

Answer 1

Step-by-step explanation:

We know that l || m and t is a transversal From the figure we know that ∠ APR and ∠ PRD are alternate angles ∠ APR = ∠ PRD We can write it as ½ ∠ APR = ½ ∠ PRD We know that PS and RQ are the bisectors of ∠ APR and ∠ PRD So we get ∠ SPR = ∠ PRQ Hence, PR intersects PS and RQ at points P and R respectively We get PS || RQ In the same way SR || PQ Therefore, PQRS is a parallelogram We know that the interior angles are supplementary ∠ BPR + ∠ PRD = 180o From the figure we know that PQ and RQ are the bisectors of ∠ BPR and ∠ PRD We can write it as 2 ∠ QPR + 2 ∠ QRP = 180o Dividing the equation by 2 ∠ QPR + ∠ QRP = 90o …… (1) Consider △ PQR Using the sum property of triangle ∠ PQR + ∠ QPR + ∠ QRP = 180o By substituting equation (1) ∠ PQR + 90o = 180o On further calculation ∠ PQR = 180o – 90o By subtraction ∠ PQR = 90o We know that PQRS is a parallelogram It can be written as ∠ PQR = ∠ PSR = 90o We know that the adjacent angles in a parallelogram are supplementary ∠ SPQ + ∠ PQR = 180o By substituting the values in above equation ∠ SPQ + 90o = 180o On further calculation ∠ SPQ = 180o – 90o By subtraction ∠ SPQ = 90o We know that all the interior angles of quadrilateral PQRS are right angles Therefore, it is proved that the quadrilateral formed by the bisectors of interior angles

Answer 2

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.