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.]

Solution:

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 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.