The figure formed by joining the mid-point of the adjacent sides of a parallelogram is a rectangle    (b)  parallelogram     (c)  rhombus    (d)  square

Explanation

Let ABCD is a parallelogram, with points E,F,G and H are the mid-points of AB, BC, CD and DA.

By mid-point theorem,

`EF=1/2AB`

Similarly,

`BF=1/2BC`

`therefore triangle BEF~triangleBAC` (BY SAS)

Now, EF is half of the length of diagonal AC since that's proportion of the similar triangles.

So, `triangleDHG~triangleDAC`

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