Given: Two lines AB and CD are parallel with transversal EF intersecting AB at M and CD at O. Bisectors of consecutive interior angles meet at N and P respectively.
To Prove: Quadrilateral MNOP is a rectangle.
Proof: Since MP, OP, MN and ON are angle bisectors
∴ ∠1 = ∠2
∠3 = ∠4
∠5 = ∠6
∠7 = ∠8
Again AB || CD and EF is a transversal
∴ ∠BMO = ∠MOC
and ∠AMO = ∠DOM (alternate angles)
∴
∠AMO =
∠DOM
∴ ∠3 = ∠7
But these are alternate angles of sides MN and OP.
∴ MN || OP
Similarly, MP || ON
Hence MNOP is a parallelogram.
Now ∠ BMO + ∠ MOD = 180° (Consececutive interior angles)
⇒
∠ BMO +
∠ MOD =
x 180° = 90°
But in ∠ MOP, we have ∠2 + ∠7 + ∠ P = 180°.
⇒
∠ BMO +
∠ MOD + ∠ P = 180°
∴ 90° + ∠ P = 180°
or ∠ P = 180° - 90° = 90°
Now since MNOP is a || gm and one angle is 90°.
Hence MNOP is a rectangle.