Question

in the figure bisector of co. interior angles intersect at p prove that angle GPH is equal to 90​

Answer 1

ANSWER

Let the angle at which the transversal intersects the lines be θ

\So, ∠BAD=θ,∠EBA=180−θ

OB bisects the ∠EBA and OA bisects the ∠BAD.

consider △AOB,∠BAO=

2

θ

∠OBA=

2

1

(180−θ)=90−

2

θ

∠BAO+∠OBA+∠BOA=180°

⟹90−

2

θ

+

2

θ

+∠AOB=180°

⟹∠AOB=90°

∴ the bisectors of internal angles on the same side of the transversal intersects at right angles.

solution