Question

$ABCD$ is a parallelogram. The diagonals $AC$ and $BD$ intersect at the point $O$. If $E,F,G\text{and}H$ are the mid-points of $AO,DO,CO\text{and}BO$ respectively, then the ratio of $(EF+FG+GH+HE)$ to $(AD+DC+CB+BA)$ is:

• A. $1:1$
• B. $1:2$
• C. $1:3$
• D. $1:4$

Answer 1

The correct option is B

$1:2$

Explanation:

Let us understand it with help of a figure.

It is given that $ABCD$ is a parallelogram $AC\text{and}BD$ intersect at $O$. $E,F,G\text{and}H$ are the mid-points of $OA,OD,OC\text{and}OB$ respectively.

In $\Delta AOD$, $E$ and $F$ are mid-points of the sides $OA\text{and}OD$ respectively.

According to the midpoint theorem - In a triangle, the line segment joining the midpoint of two sides of the triangle is half of the length of the third side.

$\Rightarrow EF=\frac{1}{2}AD$ $...\left(1\right)$

Similarly,

$FG=\frac{1}{2}DC$ $...\left(2\right)$

$GH=\frac{1}{2}CB$ $...\left(3\right)$

$HE=\frac{1}{2}BA$ $...\left(4\right)$

$$\begin{gathered} \Rightarrow EF+FG+GH+HE=\frac{1}{2}\left(AD+DC+CB+BA\right) \\ \Rightarrow \frac{EF+FG+GH+HE}{AD+DC+CB+BA}=\frac{1}{2} \end{gathered}$$

Therefore, the ratio of $\frac{EF+FG+GH+HE}{AD+DC+CB+BA}$ is $\frac{1}{2}$.

Hence, the correct option is B.