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In a Square Abcd, Its Diagonals Ac and Bd Intersect Each Other at Point O. Show That: (I) `∠`Onl + `∠`Oml = 180 (Ii)`∠`Bam = `∠`Bma (Iii) Alob is a Cyclic Quadrilateral - ICSE Class 10 - Mathematics

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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.

 

 

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Solution In a Square Abcd, Its Diagonals Ac and Bd Intersect Each Other at Point O. Show That: (I) `∠`Onl + `∠`Oml = 180 (Ii)`∠`Bam = `∠`Bma (Iii) Alob is a Cyclic Quadrilateral Concept: Cyclic Properties.
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