Question

in the given figure o q is the angular bisector of angle pqr​

Answer 1

Answer:

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