Question

$\bf \huge {\underline {\underline \red{QuEsTiOn}}}$

Two prallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.​

Answer 1

$\huge \sf {\orange {\underline {\pink {\underline { Answer᭄ \ :- }}}}}$

We know that l ∥ m and t is transversal

from the figure we know that ∠APR and ∠PRD are alternate angles

$\sf{∠APR =∠PRD}$

We can write it as

$\sf{1/2 ∠APR = 1/2 ∠PRD}$

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

$\sf{∠BPR + ∠PRD = 180°}$

from the figure we know that PQ and RQ are the bisectors of ∠BPR and ∠PRD

We can write it as

$\sf{∠QPR + ∠QRP = 90°.....(1)}$

Dividing the equation by 2

$\sf{2 ∠QPR + 2 ∠QRP = 180°}$

Consider △PQR

Using the sum property of triangle

$\sf{∠PQR + ∠QPR + ∠QRP = 180°}$

by substituting in equation (1)

$\sf{∠PQR + 90° = 180°}$

$\sf{∠PQR = 90°}$

We know that PQRS is a parallelogram

it can be written as

$\sf{∠PQR = ∠PSR = 90°}$

We know that the adjacent angles in a parallelogram are supplementary

$\sf{∠SPQ + 90° = 180°}$

$\sf{∠SPQ = 90°}$

We know that all the interior angles of quadrilateral PQRS are right angles 

therefore it is proved that the quadrilateral PQRS are right angles

$\sf\purple{➛}$therefore it is proved that the quadrilateral formed by the bisectors of interior angles is a rectangle.

$\underline{\red{\sf{\blue\:Thank \ чou ♡ }}}$

Answer 2

TO SHOW :

• The quadrilateral formed by the bisectors of interior angles is a rectangle.

PROOF :

l and m are parallel lines which are intersected by a transversal t.

l am intersects at point A and C respectively.

Bisector of interior angles intersects at B and D.

Now, the alternate angles formed are :

$\implies \sf \angle PAC \ = \ \angle ACR$

$\implies \sf \dfrac {1}{2} \ \angle PAC \ = \dfrac {1}{2} \angle ACR$

$\implies \sf \angle BAC \ = \ \angle ACD$

We know that,

The alternate angles are equal.

➠ AB || DC because the alternate angles are equal.

➠ Now, similarly BC || AD.

$\therefore$ ABCD is a parallelogram.

Here,

$\implies \sf \angle PAC + \angle CAS \ = \ 180^{\circ}$ .........{This are the linear pairs}

$\implies \sf \dfrac {1}{2} \angle PAC \ + \ \dfrac {1}{2} \angle CAS \ = \ 90^{\circ}$

$\implies \sf \angle BAC \ + \ \angle CAD \ = \ 90^{\circ}$

$\implies \sf \angle BAD \ = \ 90^{\circ}$

$\therefore$ Therefore, ABCD is a parallelogram and one of its angle is 90°.

•°• ABCD is a rectangle.