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Question

If D, E and F are respectively the midpoints of sides AB, BC and CA of ΔABC then what is the ratio of the areas of ΔDEF and ΔABC?

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Solution

Given in ΔABC, D, E and F are midpoints of sides AB, BC and CA respectively.
BC = EC
Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.
Hence DF = 1 half BC
fraction numerator D F over denominator B C end fraction 1 half → (1)
Similarly,fraction numerator D E over denominator A C end fraction = 1 half → (2)
fraction numerator E F over denominator A B end fraction = 1 half → (3)
From (1), (2) and (3) we have
fraction numerator D F over denominator B C end fraction equals fraction numerator D E over denominator A C end fraction equals fraction numerator E F over denominator A B end fraction equals fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction
But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar
Hence ΔABC ~ ΔEDF [By SSS similarity theorem]
fraction numerator a r open parentheses increment D E F close parentheses over denominator a r open parentheses increment A B C close parentheses end fraction equals fraction numerator D F squared over denominator B C squared end fraction fraction numerator D F squared over denominator B C squared end fraction equals 1 squared over 2 squared equals 1 fourth
Hence area of ΔDEF : area of ΔABC = 1 : 4

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