In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint : From D and B, draw perpendiculars to AC.]
Solution (i)
We need to show that ar(DOC)=ar(AOB)
Construction: Draw
and
In
and
, we have
(Vertically opposite angles)
(Each equal to
)
(Given)
Therefore, by AAS congruence rule, we have
(Congruent triangles are equal in area.) (1)
And,
(Corresponding parts of congruent triangles are equal.)
In
and
, we have
(Each equal to
)
(Given)
(Proved above)
Therefore, by RHS congruence rule,
(Congruent triangles are equal in area.) (2)
Adding (1) and (2), we get
Solution (ii)
We have proved above that
Adding
on both the sides, we get
Solution (iii)
We have already proved above that
(Corresponding parts of congruent triangles are equal.)
(Because
and
are alternate interior angles) (3)
It is also given that
(4)
From (3) and (4), we can say that ABCD is a parallelogram which also means that
(A quadrilateral is a parallelogram if one pair of opposite sides is equal and parallel.)
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