In the figure given below, $A B C D$ is a parallelogram whose diagonals intersect at point $O$ . $E$ is the midpoint of the side $A D$ . The side $B C$ is extended to $F$ such that $C F = \frac{1}{2} B C$ . The ratio of $D H, B H$ is

2 : 3

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1 : 3

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1 : 2

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1 : 4

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Solution

The correct option is B $1 : 3$
Here, $C F = A E$ and $C F | | A E$ .
Hence, AECF is a parallelogram $\Rightarrow F E | | A C$
Let FE intersects CD at point G
$△ D E G \sim △ D A C$ (AAA similarity)
$\frac{D H}{D O} = \frac{D E}{A D} = \frac{1}{2}$
$\Rightarrow D H = \frac{1}{2} D O = \frac{1}{4} B D$
Hence, $\frac{D H}{B H} = 1 : 3$

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