Ex 9.3, 6 - Class 9

Transcript

Ex9.3, 6 In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: (i) ar (DOC) = ar (AOB) Given: A quadrilateral ABCD where OB = OD & AB = CD To prove: ar (DOC) = ar (AOB) Proof : Let us draw DP ⊥ AC and BQ ⊥ AC. In ΔDOP and ΔBOQ, ∠DPO = ∠BQO ∠DOP = ∠BOQ OD = OB ΔDOP ≅ ΔBOQ ∴ DP = BQ & ar(DOP) = ar(BOQ) In ΔCDP and ΔABQ, ∠CPD = ∠AQB CD = AB DP = BQ ∴ ΔCDP ≅ ΔABQ ⇒ ar(CDP) = ar(ABQ) Adding (2) & (3) ar(DOP) + ar(CDP) = ar(BOQ) + ar(ABQ) ar (DOC) = ar (AOB) Hence proved Ex9.3, 6 In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: (ii) ar (DCB) = ar (ACB) In part(i) we proved that ar(DOC) = ar(AOB) Adding ar(OCB) both sides, ar(DOC) + ar(OCB) = ar(AOB) + ar(OCB) ⇒ ar(DCB) = ar(ACB) Ex 9.3, 6 In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that: (iii) DA ∥ CB or ABCD is a parallelogram. In the part(ii) we proved ar(∆DBC) = ar(∆ ABC) We know that two triangles having same base & equal areas, lie between same parallels, Here ∆DBC & ∆ABC are on the same base BC & are equal in area, So, these triangles lie between the same parallels DA and CB ∴ DA ∥ CB Now, In Δ DOA & Δ BOC ∠ DOA = ∠ BOC ∠ DAO = ∠ BCO OD = OB ∴ Δ DOA ≅ Δ BOC DA = BC So, In ABCD, since DA ∥ CB and DA = CB One pair of opposite sides of quadrilateral ABCD is equal and parallel ∴ ABCD is a parallelogram Hence proved