In Figure, 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.]

Construction: Let us draw \(DN\perp{AC}\) and \(BN\perp{AC}\).



i) In (\(\triangle{DON}\)) and (\(\triangle{BOM}\)),

\(\angle{DNO}\) = \(\angle{BMO}\) ...(By construction)
\(\angle{DON}\) = \(\angle{BOM}\)
(Vertically opposite angles)
Also, OD = OB ...(Given)
\(\therefore \) \(\triangle{DON}\) \(\displaystyle \cong\) \(\triangle{BOM}\)
(By AAS congruency test)
Thus, DN = BM ...(i)(CPCT)
But, we know that, congruent triangles have equal areas.
\(\therefore \) Area (\(\triangle{DON}\)) = Area (\(\triangle{BOM}\)) ...(ii)

Now, in (\(\triangle{DNC}\)) and (\(\triangle{BMA}\)),
\(\angle{DNC}\) = \(\angle{BMA}\) ...(By construction)
DN = BM ...(Using Equation (i))
Also, CD = AB ...(Given)
\(\therefore \) \(\triangle{DNC}\) \(\displaystyle \cong\) \(\triangle{BMA}\)
(By RHS congruency test)
\(\therefore\) Area (\(\triangle{DNC}\)) = Area (\(\triangle{BMA}\)) ...(iii)
On adding eq. (ii)and (iii), we get,
Area \(\triangle{DON}\) + Area \(\triangle{DNC}\) = Area \(\triangle{BOM}\) + Area \(\triangle{BMA}\)
\(\therefore \) Area \(\triangle{DOC}\) = Area \(\triangle{AOB}\)


ii)We have, Area \(\triangle{DOC}\) = Area \(\triangle{AOB}\)

Adding Area \(\triangle{OCB}\) on both sides, we get,

\(\Rightarrow \) Area \(\triangle{DOC}\) + Area \(\triangle{OCB}\) = Area \(\triangle{AOB}\) + Area \(\triangle{OCB}\)
\(\Rightarrow \) Area \(\triangle{DCB}\) = Area \(\triangle{ACB}\)


iii)Now, we have, Area \(\triangle{DCB}\) = Area \(\triangle{ACB}\)

We also know that, if two triangles have the same base and equal areas, then these will lie between the same parallels.

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

\(\therefore \) we can say that, ABCD is a parallelogram.
Hence, proved.