Question

In parallelogram pqrs the bisector of angle p and q meet sr at o. Show that angle poq=90

Answer 1

Answer:

Please draw the diagram by your self

Step-by-step explanation:

Given: A parallelogram With angle bisector PA and QB meeting at o

To prove: angle Poq =90  

Solution:- We know that

                Angle(QPS) +Angle(PQR)= 180 degrees ( Properties of a     parallelogram)

Now lets take angle QPA = angle 1 and Angle PQB = angle 2

There fore we can represent it as

2(Angle 1) + 2( Angle 2) = 180

so, angle 1 + angle 2 = 90.................(Equation 1)

In Triangle POQ

angle1+angle2+anglePOQ=180..................( Angle sum property )

90+angle Poq= 180............(substituting Equation 1)

angle poq=90

Hence proved

Answer 2

Step-by-step explanation:

PQRS is a parallelogram.

PO is angle bisector of ∠P

∴  ∠SPO=∠OPQ        ---- ( 1 )

QO is an angle bisector of ∠Q

∴  ∠RQO=∠OQP    ---- ( 2 )    

∴  PS∥QR

⇒  ∠SPQ+∠PQR= 180 degree    [Sum of adjacent angles are supplementary]

⇒  ∠SPO+∠OPQ+∠OQP+∠OQR=180 degree

⇒  2∠OPQ+2∠OQP=180 degree   [ From ( 1 ) and ( 2 ) ]

⇒  ∠OPQ+∠OQP=90 degree          ---- ( 3 )

Now, in △POQ,

⇒  ∠OPQ+∠OQP+∠POQ=180 degree

⇒  90 degree  +∠POQ=180 degree  [ From ( 3 ) ]

⇒  ∠POQ=90 degree

∴ LHS = RHS

Therefore, Proved

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