Question

8. Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.
If AB = CD, show that:
i) ar (AAOD) = ar (ABOC)
ii) ar (AABD) = ar (AABC)
iii) DA || CB or ABCD is a parallelogram.
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Answer 1

Answer:

Step-by-step explanation:

Given, ABCD is a quadrilateral in which AB = CD its diagonals AC and BD intersect at O such  that OB=OD.

To show:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.

Construction,

DE ⊥ AC and BF ⊥ AC are drawn.

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (each 90°)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

Therefore, ΔDOE ≅ ΔBOF

(by AAS congruence rule)

Thus, DE = BF (By CPCT) — (i)

also, ar(ΔDOE) = ar(ΔBOF) ........(ii)

(Two Congruent triangles have equal areas)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (each 90°)

CD = AB (Given)

DE = BF (From i)

Therefore,ΔDEC ≅ ΔBFA

byy RHS congruence rule)

Thus, ar(ΔDEC) = ar(ΔBFA) ........(iii)

(Two Congruent triangles have equal areas)

Adding (ii) and (iii),

ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)

 ar (DOC) = ar (AOB)

 

(ii)  ar(ΔDOC) = ar(ΔAOB)

⇒ ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)   

(Adding ar(ΔOCB) on both sides)

 ar(ΔDCB) = ar(ΔACB)  

 

(iii) From part (ii) (ΔDCB) & (ΔACB) have all areas and have the same base BC. So, (ΔDCB) & (ΔACB) must lie between the same parallels.

 DA || BC — (iv)

∠FBO= ∠EDO.....(v)

(ΔDOE ≅ ΔBOF )

∠FBA=∠EDC.....(vi)

(ΔDEC ≅ ΔBFA )

 On adding eq v & vi

∠ABD=∠CDB

Therfore, DC||AB......(vii)

From eq iv & vii, We get DA||CB & DC||AB

Hence, ABCD is parallelogram  .

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