
Given a parallelogram ABCD. Complete each statement along with the definition used:
$(i)$ $AD = $ $(ii)$ $\angle DCB = $ $(iii)$ $OC = $ $(iv)$ $m\angle DAB + m\angle CDA = $

Answer
512.4k+ views
Hint: Use the properties of parallelogram: Opposite sides are parallel and equal, opposite angles are equal and sum of adjacent angles is ${180^ \circ }$.
Complete step-by-step answer:
Given, ACBCD is a parallelogram with O as the point of intersection of its diagonals AC and BD.
$(i)$ We know that in parallelogram, lengths of opposite sides are equal. Hence, we can conclude that:
$ \Rightarrow AD = BC$
$(ii)$ We know that in parallelogram, measure of opposite angles is equal. Hence, we can conclude that:
$ \Rightarrow \angle DCB = \angle DAB$
$(iii)$ We know that in parallelogram, diagonals bisect each other. Hence, we can conclude that:
$ \Rightarrow OC = OA$
$(iv)$ We know that in parallelogram, adjacent angles are supplementary (i.e. their sum is ${180^ \circ }$). Hence, we can conclude that:
$ \Rightarrow m\angle DAB + m\angle CDA = {180^ \circ }$
Note: In parallelogram, although the diagonals bisect each other but their lengths are not equal.
In the above parallelogram, $AC \ne BD$.
Complete step-by-step answer:
Given, ACBCD is a parallelogram with O as the point of intersection of its diagonals AC and BD.
$(i)$ We know that in parallelogram, lengths of opposite sides are equal. Hence, we can conclude that:
$ \Rightarrow AD = BC$
$(ii)$ We know that in parallelogram, measure of opposite angles is equal. Hence, we can conclude that:
$ \Rightarrow \angle DCB = \angle DAB$
$(iii)$ We know that in parallelogram, diagonals bisect each other. Hence, we can conclude that:
$ \Rightarrow OC = OA$
$(iv)$ We know that in parallelogram, adjacent angles are supplementary (i.e. their sum is ${180^ \circ }$). Hence, we can conclude that:
$ \Rightarrow m\angle DAB + m\angle CDA = {180^ \circ }$
Note: In parallelogram, although the diagonals bisect each other but their lengths are not equal.
In the above parallelogram, $AC \ne BD$.
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