Question

6. 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.]
Fig. 9.25​

Answer 1

Answer:

Step-by-step explanation:

let us draw DN perpendicular to AC and BM perpendicular to AC

 (i) In ΔDON and ΔBOM,

   ∠DNO = ∠BMO ( 90°)

    ∠DON = ∠BOM (Vertically opposite angles)

   OD = OB (Given)

    By AAS congruence rule,

     ΔDON  ΔBOM

      DN = BM  (BY CPCT)   ..... (1)

We know that congruent triangles have equal areas.

Area (ΔDON) = Area (ΔBOM)  ..... (2)  

In ΔDNC and ΔBMA,

∠DNC = ∠BMA (Angles made by perpendicular)

CD = AB (Given)

DN = BM [Using equation (i)]

ΔDNC  ΔBMA (RHS congruence rule)

Area (ΔDNC) = Area (ΔBMA) (iii)

Let us draw DN AC and BM AC

On adding equations (ii) and (iii), we obtain

Area (ΔDON) + Area (ΔDNC) = Area (ΔBOM) + Area (ΔBMA)

Therefore,

Area (ΔDOC) = Area (ΔAOB)

(ii) TO prove :Area (ΔDCB) = Area (ΔACB)

we have proved,Area (ΔDOC) = Area (ΔAOB)

Area (ΔDOC) + Area (ΔOCB) = Area (ΔAOB) + Area (ΔOCB)

(Adding Area (ΔOCB) to both sides)  

Area (ΔDCB) = Area (ΔACB)

(iii)

We have proved above, Area (ΔDCB) = Area (ΔACB)  

If two triangles have the same base and equal areas, then these will lie between the same parallels

So, DA || CB (iv)

In quadrilateral ABCD, one pair of opposite sides is equal (AB = CD) and the other pair of opposite sides is parallel (DA || CB)

Therefore, ABCD is a parallelogram

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