Question

diagonal AC and BD of quadrilateral ABCD intersect at O such that OB=OD.If AB=CD, then show that:-​

Answer 1

Step-by-step explanation:

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)

rightwards double arrow space space space space space fraction numerator left parenthesis AC right parenthesis left parenthesis DE right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis AC right parenthesis left parenthesis BF right parenthesis over denominator 2 end fraction

space space space space space space space space space space space space space space space space because space space Area space of space straight a space triangle

space space space space space space space space space space space space space space space space space equals space fraction numerator Base space straight x space Corresponding space altitude over denominator 2 end fraction

rightwards double arrow space space space space DE plus BF space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis

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

ar left parenthesis increment DOC right parenthesis equals fraction numerator OC cross times DE over denominator 2 end fraction

ar left parenthesis increment AOB right parenthesis equals fraction numerator OA cross times BF over denominator 2 end fraction

∵ 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).