Question

$ABCD$ is paralellogram. The diagonal $AC$ and $BD$ intersect at a point $O$. If $E,F,G,andH$ are the mid-poinrt of $AO,DO,CO$ 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 $1:2$

ABCD is a IIgm diagonals AC and BD interest at O. E, F, G and H are the mid-points of OA, OD, OC and OB respectively.

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

$\therefore$ EF = $\frac{1}{2}AD⟶\left(1\right)[Mid-pointtheorem]$

Similarly, FG $=$ $\frac{1}{2}CD⟶\left(2\right)$

GH $=$ $\frac{1}{2}BC⟶\left(3\right)$

HE $=$ $\frac{1}{2}AB⟶\left(4\right)$

Adding, (1), (2), (3) & (4) we get

$EF+FG+GH+HE=\frac{1}{2}(AD+CD+BC+AB)$

$\Rightarrow \frac{EF+FG+GH+HE}{AD+CD+BC+BA}=\frac{1}{2}$