Show that the figure formed by joining the mid points of sides of a rhombus successively is a rectangle
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we draw the figure in which ABCD is a rhombus and E,F,G,H are respectively the mid points.
As ABCD is a rhombus so diagonals AC and BD bisect each other at 90degree so
cod=90degree.
now EF|| AC and EF=1/2AC
also HG||AC and HG=1/2AC
from above EFGH is a paralellogram.
now again from the figure POSH is a paralellogram
SOP=SPH
SOP=90degree
SPH=90degree
hence EFGH is a rectangle
As ABCD is a rhombus so diagonals AC and BD bisect each other at 90degree so
cod=90degree.
now EF|| AC and EF=1/2AC
also HG||AC and HG=1/2AC
from above EFGH is a paralellogram.
now again from the figure POSH is a paralellogram
SOP=SPH
SOP=90degree
SPH=90degree
hence EFGH is a rectangle
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