Question

In a parallelogram PQRS of the given figure the bisectors of angle B and angle Q meet SR at O. Show that angle POQ =90°

Answer 1

Proved ∠POQ = 90° if In a parallelogram PQRS bisector of ∠P & ∠Q meet SR at O

Step-by-step explanation:

in Parallelogram sum of adjacent angles = 180°

=> ∠P + ∠Q  = 180°

PO  & QO are bisector of ∠P & ∠Q

=> ∠SPO = ∠QPO = ∠P/2

& ∠RQO = ∠PQO = ∠Q/2

if we draw a line ON Parallel to PS & QR passing through

then  ∠SPO =  ∠PON

 & ∠RQO = ∠QON

∠POQ = ∠PON + ∠QON

=> ∠POQ =∠SPO  +  ∠RQO

=> ∠POQ = ∠P/2 + ∠Q/2

=> ∠POQ = (∠P + ∠Q)/2

=> ∠POQ = (180°)/2

=> ∠POQ = 90°

QED

Proved

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Answer 2

$\angle POQ = 90^\circ$

Step-by-step explanation:

here ,

in Parallelogram PQRS

we know that the opposite sides are equal and parallel

therefore ,

PS ║ QR

and PQ is a transversal line

then

$\angle P + \angle Q = 180^\circ$ (angle on the same side of the transversal )..

the bisectors of angle B and angle Q meet SR at O.

i.e

$\angle 1 = \angle 2 \\\angle 3 = \angle 4$

putting in equation 1

we get

$\angle P + \angle Q = 180^\circ$

$\angle 1 +\angle 2 +\angle 3 +\angle 4 = 180^\circ \\2 ( \angle 2 +\angle 3 ) = 180 \\\angle 2 +\angle 3 = 90^\circ$

now in triangle POQ

$\angle 2 +\angle 3 +\angle POQ = 180^\circ$

(angle sum property)

$90^\circ + \angle POQ = 180^\circ \\\angle POQ = 180-90\\\angle POQ = 90^\circ$

hence proved

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