39.

In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:

(i) ar(ΔDOC) = ar(ΔAOB)
(ii) ar(ΔDCB) = ar(ΔACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint. From D and B, draw perpendiculars to AC.]

Given: Diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB = OD. To Prove: If AB = CD, then

(i) ar(ΔDOC) = ar(ΔAOB)
(ii) ar(ΔDCB) = ar(ΔACB)
(iii) DA || CB or ABCD is a parallelogram.



Construction: Draw DE ⊥ AC and BF ⊥ AC.
Proof: (iii) In ΔADB,
∵    AO is a median
∴ ar(ΔAOD) = ar(ΔAOB)    ...(1)
∵    A median of a triangle divides it into two triangles of equal areas
In ΔCBD,
∵    CO is a median.
ar(ΔCOD) = ar(ΔCOB)    ...(2)
∵    A median of a triangle divides it into two triangles of equal areas
Adding (1) and (2), we get ar(ΔAOD) + ar(ΔCOD)
= ar(ΔAOB) + ar(ΔCOB)
⇒ ar(ΔACD) = ar(ΔACB)



In right Δs DEC and BFA,
Hyp. DC = Hyp. BA    | given
DE = BF    | From (3)
∴ ΔDEC ⊥ ABFA    | R.H.S. Rule
∴ ∠DCE = ∠BAF    | C.P.C.T.
But these angles form a pair of equal alternate interior angles.
∴ DC || AB    ...(4)
∵ DC = AB and DC || AB ∴ □ABCD is a parallelogram.
∵ A quadrilateral is a parallelogram if a pair of opposite sides are parallel and equal DA || CB
| ∵ Opposite sides of a || gm are parallel (i) ∵ ABCD is a parallelogram
∴ OC = OA    ...(5)
Diagonals of a parallelogram bisect each other

∵ DE = BF    | From (3)

and    OC = OA    | From (5)

∴ ar(ΔDOC) = ar(ΔAOB).

(ii) From (i),
ar(ΔDOC) = ar(ΔAOB)
Δ ar(ΔDOC) + ar(ΔOCB)
= ar(ΔAOB) + ar(ΔOCB)
| Adding equal areas on both sides ⇒ ar(ΔDCB) = ar(ΔACB).