#### 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:

(i) `∠`ONL + `∠`OML = 180

(ii)`∠`BAM = `∠`BMA

(iii) ALOB is a cyclic quadrilateral

#### Solution

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

i)

∴ `∠`AOB = `∠`AOD = 90°

In Δ ANB ,

`∠`ANB = 180° - (`∠`NAB + `∠`NBA)

⇒ `∠`ANB = 180° - `(45° + 45°/2)`(NB is bisector of `∠`ABD)

⇒ `∠`ANB = 180° - 45° - `(45°)/2 = 135° - (45°)/2`

But, `∠`LNO = ∠`ANB (vertically opposite angles)

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

Now in ΔAMO,

`∠`AMO = 180° - ( `∠`AOM + `∠`OAM)

⇒ `∠`AMO = 180° - `(90 + (45°)/2)` (MA is bisector of `∠`DAO)

⇒ `∠` AMO = 180° - 90° - `(45°)/2 = 90° - (45°)/2` ...................(ii)

Adding (i) and (ii)

`∠`LNO + `∠`AMO = 135 - `(45°)/2 + 90° - (45°)/2`

⇒ `∠`LNO + `∠`AMO = 225° - 45° = 180°

⇒ `∠`ONL + `∠`OML = 180°

ii)

`∠`BAM = `∠`BAO + `∠`OAM

⇒ `∠BAM = 45° + (45°)/2 = 67 (1°)/2`

And

⇒ `∠`BMA = 180° - (`∠`AOM + `∠`OAM)

⇒ `∠`BMA = 180° - 90° - `(45°)/2 = 90° - (45°)/2 = 67(1°)/2`

∴ `∠`BAM = `∠`BMA

iii) In quadrilateral ALOB,

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

Therefore, ALOB is a cyclic quadrilateral.