In 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)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
Given,
OB = OD and AB = CD
Construction,
DE ⊥ AC and BF ⊥ AC are drawn.
Proof:
(i) In ΔDOE and ΔBOF,
∠DEO = ∠BFO (Perpendiculars)
∠DOE = ∠BOF (Vertically opposite angles)
OD = OB (Given)
∴ ΔDOE ≅ ΔBOF by AAS congruence condition.
∴ DE = BF (By CPCT) — (i)
ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) — (ii)
In ΔDEC and ΔBFA,
∠DEC = ∠BFA (Perpendiculars)
CD = AB (Given)
DE = BF (From (i))
∴ ΔDEC ≅ ΔBFA by RHS congruence condition.
∴ ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) — (iii)
Adding (ii) and (iii),
ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)
⇒ ar (DOC) = ar (AOB)
(ii) ar(ΔDOC) = ar(ΔAOB)
Adding ar(ΔOCB) in LHS and RHS, we get,
⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)
⇒ ar(ΔDCB) = ar(ΔACB)
(iii) When two triangles have same base and equal areas, the triangles will be in between the same parallel lines
ar(ΔDCB) = ar(ΔACB)
DA || BC — (iv)
For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel.
∴, ABCD is parallelogram.