Question

ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points 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

(b) 1 : 2

By mid-point theorem, in ∆s ODA, OCD, OBC, OBA respectively

\(\frac{EF}{AD} = \frac{FG}{DC}=\frac{GH}{CB}=\frac{HE}{BA}=\frac{1}{2}\)

If \(\frac{a}{b}=\frac{c}{b}=\frac{e}{f}=.............,\)

then each ratio is equal to \(\frac{a+b+c+e+.......}{b+d+f+........}\)

⇒ \(\frac{(EF+FG+GH+HE)}{(AD+DC+CB+BA)}\) = \(\frac{1}{2}.\)