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Mathematics

In the given figure, D, E and F are the mid-points of the sides BC, AC and AB respectively of ΔABC. Then which of the following does not hold true?

  1. ΔAFE ∼ ΔABC

  2. ΔFBD ∼ ΔABC

  3. ΔEDC ∼ ΔABC

  4. ΔDFE ∼ ΔABC

In the given figure, D, E and F are the mid-points of the sides BC, AC and AB respectively of ΔABC. Then which of the following does not hold true? Similarity of Triangles, RSA Mathematics Solutions ICSE Class 10.

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Answer

Given,

Since D, E, F are midpoints of BC, AC and AB respectively.

According to the Mid-point Theorem, the segment joining the mid-points of two sides of a triangle is parallel to the third side and is half its length.

Thus, by mid-point theorem,

DE ∥ AB

DE = 12\dfrac{1}{2}AB

EF ∥ BC

EF = 12\dfrac{1}{2}BC

FD ∥ AC

DF = 12\dfrac{1}{2}AC

Ratios can be written as,

⇒DFAC=EFBC=DEAB⇒12ACAC=12BCBC=12ABAB⇒12=12=12\Rightarrow \dfrac{DF}{AC} = \dfrac{EF}{BC} = \dfrac{DE}{AB} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}AC}{AC} = \dfrac{\dfrac{1}{2}BC}{BC} = \dfrac{\dfrac{1}{2}AB}{AB} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2}

∴ ΔDFE ∼ ΔACB is true, but ΔDFE ≁ ΔABC.

Hence, option 4 is the correct option.

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