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 perpendicular to AC and BM perpendicular to 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 (i)
We know that congruent triangles have equal areas.
Area (ΔDON) = Area (ΔBOM) (ii)
In ΔDNC and ΔBMA,
DNC = BMA (By construction)
CD = AB (Given)
DN = BM [Using equation (i)]
ΔDNC ΔBMA (RHS congruence rule)
Area (ΔDNC) = Area (ΔBMA) (iii)
Let us draw DN AC and BM AC
On adding equations (ii) and (iii), we obtain
Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)
Therefore,
Area (ΔDOC) = Area (ΔAOB)
(ii) Area (ΔDOC) = Area (ΔAOB)
Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)
(Adding Area (ΔOCB) to both sides)
Area (ΔDCB) = Area (ΔACB)
(iii) Area (ΔDCB) = Area (ΔACB)
If two triangles have the same base and equal areas, then these will lie between the same parallels
DA || CB (iv)
In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of opposite sides is parallel (DA || CB)
Therefore, ABCD is a parallelogram