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)
also, ar(ΔDOE) =ar(ΔBOF) (Congruent triangles) — (ii)
Now,
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 LHSand RHS, we get,
⇒ar(ΔDOC) + ar(ΔOCB) =ar(ΔAOB) + ar(ΔOCB)
⇒ ar(ΔDCB) =ar(ΔACB)
(iii) When twotriangles have same base and equal areas, the triangles will be in between thesame 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 sidesare parallel.
∴, ABCD isparallelogram.