      1. Chapter 8 Class 9 Quadrilaterals
2. Serial order wise
3. Ex 8.1

Transcript

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 ABC and DCB, AB = DC ABC = DCB BC = BC ABC DCB 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 AOB and COD, AOB = COD ABO = CDO AB = CD AOB COD AO = CO and OB = OD Hence, the diagonals of a square bisect each other. In AOB and COB, OA = OC AB = BC BO = BO AOB COB 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 ABC and DCB, AB = DC ABC = DCB BC = BC ABC DCB ABC DCB 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 AOB and COB, OA = OC AB = BC BO = BO AOB COB 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 