Question

# In the given 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) $$DA\parallel CB$$ or $$ABCD$$ is a parallelogram.

Solution

## Construction: Draw $$DE \bot AC$$ and $$BF \bot AC$$.In $$\triangle DEO$$ and $$\triangle BFO$$$$\angle DEO= \angle BFO$$          ....By construction $$\angle DOE= \angle BOF$$          ....Vertically opposite angles $$OD = OB$$                     .....Given$$\triangle DOE \cong \triangle BOF$$     ...SAA test of congruence    ....(1)$$\therefore DE=BF$$          .... CSCT        ....(2)and $$OE = OF$$            ... CSCT  ....(3)In $$\triangle DCE$$ and $$\triangle BFA$$ $$\angle DEC= \angle BFA$$          ..... By construction$$DC = AB$$         ....Given$$DE = BF$$          ....From (2)$$\triangle DEC \cong \triangle BFA$$    .....(RHS) Right Hypotenuse Side test    ....(4)$$\therefore EC = AF$$       ......C.S.C.T.      ....(5)Similarly, we can prove $$\triangle DOA \cong \triangle BOC$$    ....(6)Adding (1) and (2), we get $$OE+EC = OF+AF$$$$\therefore OC = OA$$    .....(7)(i) Adding (1) & (4), we get,$$A(\triangle DOE) + A (\triangle DEC) = A (\triangle BOF) + A (\triangle BFA)$$$$\therefore A(\triangle DOC) = A (\triangle BOA)$$      ....(8)(ii) Adding (6) and (8), we get$$A (\triangle DOC) + A (\triangle DOA) = A (\triangle BOA) + A (\triangle BOC)$$$$\Rightarrow A (\triangle ACD) = A (\triangle ACB)$$(iii) So, from (7) and given,$$\Box ABCD$$ is a parallelogram       .....(Diagonals bisect each other)and $$DA \parallel CB$$      ....Opp. sides of parallelogram are parallel to each otherMathematicsRS AgarwalStandard IX

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