# What is the ratio of (EF+FG+GH+HE) and (AD+DC+CB+BA)?

ABCD is a parallelogram and the diagonals AC and BD intersect at ‘O’. If E, F, G and H are the midpoints of AO, DO, CO and BO respectively, then what is the ratio of (EF+FG+GH+HE) and (AD+DC+CB+BA)?

- 1:2
- 2:3
- 3:4
- 4:5

**Answer: (1) 1: 2**

**Solution:-**

According to the question,

In the diagram-

In triangle AOD,

AD II EF and F , F are the midpoint of OA and OD

Then EF = AD/2

Similarly, FG =AB/2, HG = BC/2 & EH = CD/2

Then (EF + FG + GH + HE ) = (AD + DC + CB + BA)/2

(EF + FG + GH + HE)/ (AD + DC + CB + BA) = 1/2

So, the ratio of (EF + FG + GH + HE) and (AD + DC + CB + BA) is 1: 2.

Hence, the correct answer is option (1) 1: 2.