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Question

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.
1715709_427c1c6396cb403dbc97e555afac22e9.PNG

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Solution

We know that lm and t is transversal
from the figure we know that APR and PRD are alternate angles
APR=PRD
We can write it as
1/2APR=1/2PRD
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
PSRQ
in the same way SRPQ
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
2QPR+2QRP=180o
Dividing the equation by 2
QPR+QRP=90o.....(1)
Consider PQR
Using the sum property of triangle
PQR+QPR+QRP=180o
by substituting in equation (1)
PQR+90o=180o
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+90o=180o
SPQ=90o
We know that all the interior angles of quadrilateral PQRS are right angles
therefore it is proved that the quadrilateral PQRS are right angles
therefore it is proved that the quadrilateral formed by the bisectors of interior angles is a rectangle.

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