NCERT Solutions
Class 9 Maths
Areas of Parallelograms and Triangles
Ex.9.3 Q.6
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.]
Let us draw DN ⊥ AC and BM ⊥ AC.
1. In ∆DON and ∆BOM,
∠DNO = ∠BMO (By construction)
∠DON = ∠BOM (Vertically opposite angles)
OD = OB (Given)
By AAS congruence rule,
∆DON ≅ ∆BOM
DN = BM ... (1)
We know that congruent triangles have equal areas.
Area (∆DON) = Area (∆BOM) ... (2)
In ∆DNC and ∆BMA,
∠DNC = ∠BMA (By construction)
CD = AB (Given)
DN = BM
[Using equation (1)]
So, ∆DNC ≅ ∆BMA
(RHS congruence rule)
Area (∆DNC) = Area (∆BMA) ... (3)
On adding equations (2) and (3), we obtain
Area (∆DON) + Area (∆DNC) = Area (∆BOM) + Area (∆BMA)
Therefore, Area (∆DOC) = Area (∆AOB)
(ii) We obtained,
Area (∆DOC) = Area (∆AOB)
Area (∆DOC) + Area (∆OCB) = Area (∆AOB) + Area (∆OCB)
[Adding Area (∆OCB) to both sides]
Area (∆DCB) = Area (∆ACB)
(iii) We obtained,
Area (∆DCB) = Area (∆ACB)
If two triangles have the same base and equal areas, then these will lie between the same
parallels.
So, DA || CB ... (4)
In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of
opposite sides are parallel (DA || CB).
Therefore, ABCD is a parallelogram.