prove that area of triangle formed by joining the mid points of the sides of a triangle is equal to one fourth of triangle
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Answer:
Let DEF be the midpoints of sides of a triangle ABC( with D on BC, E on AB and F on AC ). Now, considering triangles AEF and ABC, angles EAF = BAC and AE / AB = 1/2 and AF/AC = 1/2.
Hence, both triangles are similar by the SAS ( Side - Angle - Side ) criterion and correspondingly as AE/AB=AF/AC=EF/BC ( similar triangle properties ), EF =BC/2.
the cases DF=AC/2 and DE=AB/2 can be proved in the same way. Now, triangle DEF and ABC are similar with DE/AB=EF/BC=DF/AC =1/2. Similar triangles have the property :
Ratio of areas = square of ratios of sides
and hence, ( area of triangle DEF )/( area of triangle ABC ) = 1/2*1/2 = 1/4.
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