Question

plz solve this question immediately.

Answer 1

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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Hope this helps!!!