Text Solution
Solution : Given : A quadrialteral ABCD whose diagonals AC and BD bisect each other at right angles. <br> `i.e.," "OA=OCandOB=OD` <br> `and" "angleAOD=angleAOB=angleCOD=angleBOC=90^(@)` <br> <img src="//d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_IX_C08_S01_030_S01.png" width="80%"> <br> To prove : ABCD is a rhombus. <br> Proof. In `DeltaOAB and DeltaODC,` we have <br> `because{:{(OA=OC,("given")),(OB=OD,("given")),(angleAOB=angleCOD,("vertically opposite angles")):}` <br> `therefore" "DeltaOAB~=DeltaOCD" "("by SAS rule")` <br> `therefore" "AB=CD" "("c.p.c.t")...(1)` <br> `because{:{(OA=OC,("given")),(OD=OB,("given")),(angleAOD=angleBOC,("vertically oposite angles")):}` <br> `therefore" "DeltaOAD~=DeltaOCB" "("by SAS rule")` <br> `therefore" "AD=BC" "(c.p.c.t.)...(2)` <br> Similarly, we can prove that <br> `" "{:(AB=AD),(CD=BC):}" "...(3)` <br> Hence, from (1), (2) and (3), we get <br> `" "AB=BC=AD=CD` <br> Hence, ABCD is a rhombus.
Last updated at Jan. 31, 2022 by
This video is only available for Teachoo black users
This video is only available for Teachoo black users
Solve all your doubts with Teachoo Black (new monthly pack available now!)
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 AOD and COB, OA = OC AOD = COB OD = OB AOD COB 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 AOD and COD, OA = OC AOD = COD OD = OD AOD COD 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 AOD and COD, OA = OC AOD = COD OD = OD AOD COD 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: Since, 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. Given, 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, we have to prove all the sides of ABCD are equal. Now, In ΔAOD and ΔCOD, OA = OC ∠AOD = ∠COD OD = OD ∴ ΔAOD ≅ ΔCOD So, AD = CD Similarly, we can prove that AD = AB and AB = BC Thus, AB = CD = AD = BC Since, all sides of parallelogram ABCD is equal ∴ ABCD is a rhombus Hence proved