# If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.

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.

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