diagonals ACand BD of quadrilateral ABCD intersect at O such that OB = OD. if AB = CD , then show that :

1.ar(DOC) = ar(AOB).

2.ar(DCB) = ar(ACB)

3.DA ll CB or ABCD is a parallelogram.

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**Let us draw DN ⊥ AC and BM ⊥ AC.**

**(i) In ΔDON and ΔBOM,**

**∠DNO = ∠BMO (By construction)**

**∠DON = ∠BOM (Vertically opposite angles)**

**OD = OB (Given)**

**By AAS congruence rule,**

**ΔDON ≅ ΔBOM**

**∴ DN = BM ... (1)**

**We know that congruent triangles have equal areas.**

**∴ Area (ΔDON) = Area (ΔBOM) ... (2)**

**In ΔDNC and ΔBMA,**

**∠DNC = ∠BMA (By construction)**

**CD = AB (Given)**

**DN = BM [Using equation (1)]**

**∴ ΔDNC ≅ ΔBMA (RHS congruence rule)**

**⇒ Area (ΔDNC) = Area (ΔBMA) ... (3)**

**On adding equations (2) and (3), we obtain**

**Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)**

**Therefore, Area (ΔDOC) = Area (ΔAOB)**

**(ii) We obtained,**

**Area (ΔDOC) = Area (ΔAOB)**

**⇒ Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)**

**(Adding Area (ΔOCB) to both sides)**

**⇒ Area (ΔDCB) = Area (ΔACB)**

**(iii) We obtained,**

**Area (ΔDCB) = Area (ΔACB)**

**If two triangles have the same base and equal areas, then these will lie between the same parallels.**

**∴ DA || CB ... (4)**

**In ΔDOA and ΔBOC,**

**∠DOA = ∠BOC (Vertically opposite angles)**

**OD = OB (Given)**

**∠ODA = ∠OBC (Alternate opposite angles)**

**By ASA congruence rule,**

**ΔDOA ≅ ΔBOC**

**∴ DA = BC ... (5)**

**In quadrilateral ABCD, one pair of opposite sides is equal and parallel (AD = BC)**

**Therefore, ABCD is a parallelogram.**

**:) :)**

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