In Fig. 9.25, 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) 

(ii) ar (DCB) = ar (ACB)

 (iii) DA || CB or ABCD is a parallelogram.

 [Hint: From D and B, draw perpendiculars to AC.

We can draw a perpendicular from vertices B and D on diagonal AC which will help us to make congruent triangles and we know that congruent triangles are always equal in areas

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

Let us construct DN ⊥ AC and BM ⊥ AC.

i) In ∆DON and ∆BOM,

∠DNO = ∠BMO = 90° (By construction)

∠DON = ∠BOM (Vertically opposite angles are equal)

OD = OB (Given)

By AAS congruency rule, ΔDON ≅ ΔBOM

DN = BM    (By CPCT) ... (1)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM) ...(2)

Now, In ΔDNC and ΔBMA,

∠DNC = ∠BMA = 90° (By construction)

CD = AB (given)

DN = BM [Using Equation (1)]

ΔDNC ≅ ΔBMA 

Area (ΔDNC) = Area (ΔBMA ) ... (3)

On adding Equations (2) and (3), we obtain

ar (ΔDON) + ar (ΔDNC) = ar (ΔBOM) + ar (ΔBMA)

Therefore, Area (ΔDOC) = Area (ΔAOB)