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Ex 8.1, 3 - Show that if diagonals of a quadrilateral bisect - Ex 8.1


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  1. Chapter 8 Class 9 Quadrilaterals
  2. Serial order wise
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Ex .8.1,3 (Method 1) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Given: Let ABCD be a quadrilateral, where diagonals bisect each other ∴ OA = OC, and OB = OD, And they bisect at right angles So, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° To prove :ABCD a rhombus, Proof : Rhombus is a parallelogram with all sides equal We will first prove ABCD is a parallelogram and then prove all the sides of ABCD are equal. In ΔAOD and ΔCOB, OA = OC ∠AOD = ∠COB OD = OB ∴ ΔAOD ≅ ΔCOB ⇒ ∠OAD = ∠ OCB For sides AD & BC with transversal AC, ∠OAD & ∠ OCB are alternate angles, and they are equal, So, AD ∥ BC Similarly, AB ∥ DC Now, In ABCD, AD ∥ BC & AB ∥ DC Since opposites sides of ABCD are parallel, ⇒ ABCD is a parallelogram Now, we need to prove ABCD is a rhombus, i.e. all sides equal In ΔAOD and ΔCOD, OA = OC ∠AOD = ∠COD OD = OD ∴ ΔAOD ≅ ΔCOD AD = CD But AD = CB & CD = AB From (4) & (5) ⇒ AD = CD = CB = AB In ABCD, all sides are equal and it is a parallelogram. So, ABCD is a rhombus Ex .8.1,3 (Method 2) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Given: Let ABCD be a quadrilateral, where diagonals bisect each other ∴ OA = OC, and OB = OD, And they bisect at right angles So, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° To prove :ABCD a rhombus, Proof : Rhombus is a parallelogram with all sides equal We will first prove ABCD is a parallelogram and then prove all the sides of ABCD are equal. In ΔAOD and ΔCOD, OA = OC ∠AOD = ∠COD OD = OD ∴ ΔAOD ≅ ΔCOD AD = CD Similarly, we can prove that AD = AB & AB = BC From (4) & (5) ⇒ AD = CD = AB = BC In ABCD, AB = CD & AD = BC Both pairs of opposite sides are equal So, ABCD is a parallelogram Also, AB = CD = AD = BC All sides of parallelogram ABCD is equal ∴ ABCD is a rhombus Ex .8.1,3 (Method 3) Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus. Given: Let ABCD be a quadrilateral, where diagonals bisect each other ∴ OA = OC, and OB = OD, And they bisect at right angles So, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° To prove :ABCD a rhombus, Proof : Rhombus is a parallelogram with all sides equal We will first prove ABCD is a parallelogram and then prove all the sides of ABCD are equal. Since Diagonals bisect each other, Using Theorem 8.7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram So, ABCD is a parallelogram Now, In ΔAOD and ΔCOD, OA = OC ∠AOD = ∠COD OD = OD ∴ ΔAOD ≅ ΔCOD AD = CD Similarly, we can prove that AD = AB & AB = BC Thus, AB = CD = AD = BC All sides of parallelogram ABCD is equal ∴ ABCD is a rhombus Hence proved

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