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Ex 8.1, 4 - Show that diagonals of a square are equal - Ex 8.1

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  1. Chapter 8 Class 9 Quadrilaterals
  2. Serial order wise
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Ex 8.1, 4 Show that the diagonals of a square are equal and bisect each other at right angles. Given: ABCD be a square. Diagonals intersect at O. To prove : We need to prove 3 things 1. The diagonals of a square are equal ,i.e. AC = BD 2. bisect each other, i.e. OA = OC & OB = OD, 3. at right angles ,any of ∠AOB , ∠BOC , ∠COD , ∠AOD is 90° Proof: In ΔABC and ΔDCB, AB = DC ∠ABC = ∠DCB BC = BC ∴ ΔABC ≅ ΔDCB ⇒ AC = DB Hence, the diagonals of a square are equal in length. Now we need to prove diagonals bisect each other i.e. AO = CO, BO = DO In ΔAOB and ΔCOD, ∠AOB = ∠COD ∠ABO = ∠CDO AB = CD ∴ ΔAOB ≅ ΔCOD ∴ AO = CO and OB = OD Hence, the diagonals of a square bisect each other. In ΔAOB and ΔCOB, OA = OC AB = BC BO = BO ∴ ΔAOB ≅ ΔCOB ∴ ∠AOB = ∠COB Now ∠AOB + ∠COB = 180° ∠AOB + ∠AOB = 180° 2∠AOB = 180° ∠AOB = (180° )/2 ∠AOB = 90° Hence, AC & BD bisect at right angles Hence proved Ex 8.1, 4 (Method 2) Show that the diagonals of a square are equal and bisect each other at right angles. Given: ABCD be a square. Diagonals intersect at O. To prove : We need to prove 3 things 1. The diagonals of a square are equal ,i.e. AC = BD 2. bisect each other, i.e. OA = OC & OB = OD, 3. at right angles ,any of ∠AOB , ∠BOC , ∠COD , ∠AOD is 90° Proof: In ΔABC and ΔDCB, AB = DC ∠ABC = ∠DCB BC = BC ∴ ΔABC ≅ ΔDCB ΔABC ≅ ΔDCB ⇒ AC = DB Hence, the diagonals of a square are equal in length. Now we need to prove diagonals bisect each other i.e. AO = CO, BO = DO Since Square is a parallelogram & Diagonals of a parallelogram bisect each other Hence, the diagonals of a square bisect each other. Hence, OA = OC & OB = OD In ΔAOB and ΔCOB, OA = OC AB = BC BO = BO ∴ ΔAOB ≅ ΔCOB ∴ ∠AOB = ∠COB Now ∠AOB + ∠COB = 180° ∠AOB + ∠AOB = 180° 2∠AOB = 180° ∠AOB = 180° ﷮2﷯ ∠AOB = 90° Hence, AC & BD bisect at right angles Hence proved

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