is CD a diagonal in the given figure?give reason
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Answers
Answer:
Step-by-step explanation:
Hi
Parallelogram :
A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.
Two Triangles having the same base and equal areas lie between the same parallels.
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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
CD is not a diagonal in the given figure because CD is parallel to AB.
Note : A diagonal cannot be parallel to any other side
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