In parallelogram ABCD, AB = 2 AD; Let us prove that, the bisectors of ∠BAD and ∠ABC meet at the mid-point of the side DC in right angle.

Given: A parallelogram ABCD, such that AB = 2AD, let EF be perpendicular of ∠BAD and GH be perpendicular bisector of ∠ABC.

To prove: The bisectors of ∠BAD and ∠ABC meet at the mid-point of the side DC in right angle i.e. ∠AOB = 90° and O lies on CD and O is mid-point of CD.

As, EF is the bisector of ∠BAD

……[1]

Also, GH is bisector of ∠ABC

……[2]

Adding [1] and [2], we get

Also, AD || BC

⇒ ∠BAD + ∠ABC = 180° [Sum of interior angles on the same side of transversal is 180°]

……[3]

Also, In ΔAOB, By angle sum property

∠OAB + ∠AOB + ∠OBA = 180°

⇒ 90° + ∠AOB = 180°

⇒ ∠AOB = 90°

In ΔOAD and ΔOCB, By angle sum property

∠OAD + ∠ADO + ∠AOD = 180°

⇒ ∠OAB + ∠ADO +∠AOD= 180° ……[4] [∵ from [1], ∠OAB = ∠OAD]

∠OBC + ∠BOC + ∠OCB = 180°

⇒ ∠OBA + ∠BOC +∠OCB= 180° ……[5] [∵from [2], ∠OBC = ∠OBA]

Adding [4] and [5]

∠OAB + ∠ADO + ∠AOD + ∠OBA + ∠BOC + ∠OCB = 180° + 180°

⇒ (∠OAB + ∠OBA) + (∠ADO + ∠BCO) + (∠AOD + ∠BOC) = 180°

Now,

[∠OAB + ∠OBA = 180°, From 3 and ∠ADO + ∠BCO = 180°, Interior angles on same side]

⇒ 90° + 180°+ ∠AOD + ∠BOC = 360°

⇒ ∠AOD + ∠BOC = 90°

⇒ ∠AOD + ∠AOB + ∠BOC = 90 + 90 = 180°

This becomes a linear pair

Hence, DOC is a straight line i.e. O lies on CD.

Now,

∠OAB = ∠AOD [Alternate interior angles]

⇒ ∠OAD = ∠AOD [From 1]

⇒ AD = OD [Sides opposite to equal angles are equal]

Now, Given AB = 2AD and As AB = CD [opposite sides of a parallelogram are equal]

⇒ O is the mid-point of CD

Hence Proved!