Question

In the given figure, if $E$ and $F$ are the midpoints of $AB$ and $CD$ of parallelogram $ABCD$, which one is true?

• A. $CE$ trisects $BD$
• B. $AF$ trisects $BD$
• C. $\Delta ADF\cong \Delta CBE$
• D. All of these

Answer 1

Answer: (D)

All of these

In a parallelogram $ABCD$, $E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively.

Therefore, $AB\parallel DC$ ( as opposite sides of a parallelogram $ABCD$)

$AE\parallel FC$ ...........(1)

$\Rightarrow AB=DC$( as opposite sides of a parallelogram $ABCD$)

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}DC$ ....(half of equals are equal)

$\Rightarrow$$AE=CF$..........(2)

From equations (1) and (2), $AECF$ is a parallelogram.

Therefore, $EC\parallel AF$..........(3) ....( as opposite sides of parallelogram $AECF$)

In $\Delta DBC$, $F$ is the midpoint of $DC$ and $FP\parallel CQ$ ( as $EC\parallel AF$)

Therefore, $P$ is the midpoint of $DQ$ ( by converse of mid point theorem) which implies $DP=PQ$.

Similarly, in $\Delta BAP$, $BQ=PQ$

From equations (4) and (5), we obtain $DP=PQ=BQ$ which implies line segments $AF$ and $EC$ trisect the diagonal $BD$.

Hence, option D is correct.