Option : (B)
Let ABCD be a quadrilateral, with points
E, F, G and H the midpoints of
AB, BC, CD, DA respectively.
(I suggest you draw this, and add segments EF, FG, GH, and HE, along with diagonals AC and BD)
EF = \(\frac{1}{2}\)AB (Definition of midpoint)
Similarly,
BF =\(\frac{1}{2}\)BC
Thus,
Triangle BEF is similar to triangle BAC (SAS similarity)
Therefore EF is half the length of diagonal AC,
Since that's the proportion of the similar triangles.
Similarly,
We can show that triangle DHG is similar to triangle DAC,
Therefore,
HG is half the length of diagonal AC
So,
EF = HG
Similarly,
We can use similar triangles to prove that EH and FG are both half the length of diagonal BD, and therefore equal
This means that both pairs of opposite sides of quadrilateral EFGH are equal, so it is a parallelogram.
