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# 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.