Question

In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L. Show that:

1. ∠ONL + ∠OML = 180°
2. ∠BAM + ∠BMA
3. ALOB is a cyclic quadrilateral.

Answer 1

ABCD is a square whose diagonals AC and BD intersect each other at right angles at O.

i. ∴ ∠AOB = ∠AOD = 90°

In ΔANB,

∠ANB = 180° – (∠NAB + ∠NBA)

`=>` `∠ANB = 180^circ - (45^circ + (45^circ)/2)`  ...(NB is bisector of ∠ABD)

`=> ∠ANB = 180^circ - 45^circ - 45^circ/2 = 135^circ - 45^circ/2`

But, ∠LNO = ∠ANB  ...(Vertically opposite angles)

∴ `∠LNO = 135^circ - 45^circ/2`   ...(i)

Now in ΔAMO,

∠AMO = 180° – (∠AOM + ∠OAM)

`=> ∠AMO = 180^circ - (90^circ + (45^circ)/2)`  ...(MA is bisector of ∠DAO)

`=>∠AMO = 180^circ - 90^circ - (45^circ)/2 = 90^circ - (45^circ)/2`  ...(ii)

Adding (i) and (ii)

`∠LNO  + ∠AMO = 135^circ - 45^circ/2 + 90^circ - 45^circ/2`

`=>` ∠LNO + ∠AMO  = 225° – 45° = 180°

`=>` ∠ONL + ∠OML = 180°

ii. ∠BAM = ∠BAO + ∠OAM

`=> ∠BAM = 45^circ + (45^circ)/2 = 67 1^circ/2`

And

`=>` ∠BMA = 180° – (∠AOM + ∠OAM)

`=> ∠BMA = 180^circ - 90^circ - 45^circ/2`

= `90^circ - 45^circ/2`

=  `67 1^circ/2`

∴ ∠BAM = ∠BMA

iii. In quadrilateral ALOB,

∵ ∠ABO + ∠ALO = 45° + 90° + 45° = 180°

Therefore, ALOB is a cyclic quadrilateral.