Answer

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Hint: Consider the properties of parallelogram (we know that $AB\parallel CD\text{ and }AD\parallel BC$ for a parallelogram) to solve the question.

From the figure, consider the parallelogram ABCD.

From the properties of parallelogram, we know that $AB\parallel CD\text{ and }AD\parallel BC$

i.e., $AB$ is parallel to $CD$ and

$AD$ is parallel to $BC$

Which can also be considered as sides $AB$ and$DC$ are equal

$AB=DC$

Similarly $AD=BC$

i.e., Both pairs of opposite sides are parallel and they are congruent.

From the figure, it's clear that E is the midpoint of side \[AB.\]

i.e. $AE=EB$

It’s also given that $EC$ is perpendicular to $ED$ and they form an angle of $90{}^\circ $.

i.e., $\angle DEC=90{}^\circ $

In the case of parallelogram $ABCD,\text{ }\angle A=\angle C\text{ and }\angle B=\angle D$ .

From the figure we can find that $ED\ne EC.$ i.e., they are not of the same length.

Which means both $ED\text{ and }EC$are greater than the length $DC$

$\Rightarrow ED>DC\text{ and E}C>DC$

Now adding them together

\[\frac{\begin{align}

& ED>DC \\

& ED>DC \\

\end{align}}{ED+EC>2DC}\]

Let us consider that $BC$ is greater than $DC$

$\therefore $ Equation becomes $\Rightarrow ED+EC=2BC.$

So option D is correct.

Note: Remember the properties of parallelogram, with which we have to solve this equation.

As $EC\bot ED,$ students may miscalculate that $EC=ED$ which states that option A is wrong.

From the figure, $EA\ne ED$ i.e., $EA$ is shorter than the length of $ED,$ So option C is wrong.

From the figure, consider the parallelogram ABCD.

From the properties of parallelogram, we know that $AB\parallel CD\text{ and }AD\parallel BC$

i.e., $AB$ is parallel to $CD$ and

$AD$ is parallel to $BC$

Which can also be considered as sides $AB$ and$DC$ are equal

$AB=DC$

Similarly $AD=BC$

i.e., Both pairs of opposite sides are parallel and they are congruent.

From the figure, it's clear that E is the midpoint of side \[AB.\]

i.e. $AE=EB$

It’s also given that $EC$ is perpendicular to $ED$ and they form an angle of $90{}^\circ $.

i.e., $\angle DEC=90{}^\circ $

In the case of parallelogram $ABCD,\text{ }\angle A=\angle C\text{ and }\angle B=\angle D$ .

From the figure we can find that $ED\ne EC.$ i.e., they are not of the same length.

Which means both $ED\text{ and }EC$are greater than the length $DC$

$\Rightarrow ED>DC\text{ and E}C>DC$

Now adding them together

\[\frac{\begin{align}

& ED>DC \\

& ED>DC \\

\end{align}}{ED+EC>2DC}\]

Let us consider that $BC$ is greater than $DC$

$\therefore $ Equation becomes $\Rightarrow ED+EC=2BC.$

So option D is correct.

Note: Remember the properties of parallelogram, with which we have to solve this equation.

As $EC\bot ED,$ students may miscalculate that $EC=ED$ which states that option A is wrong.

From the figure, $EA\ne ED$ i.e., $EA$ is shorter than the length of $ED,$ So option C is wrong.

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