CBSE Class 10 Mathematics Standard sample paper 2024-25: question 26(A)

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Step-by-step solution

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1. Given: $D,E,F$ are the midpoints of $BC,CA,AB$ of $△ABC$. To prove: $△FBD\sim △DEF$ and $△DEF\sim △ABC$.

2. By the midpoint theorem (the segment joining the midpoints of two sides is parallel to the third side and half of it):
   $FE\parallel BC$, $FE=\frac{1}{2}BC$; $DE\parallel AB$, $DE=\frac{1}{2}AB$; $FD\parallel AC$, $FD=\frac{1}{2}AC$.
   Also $FB=\frac{1}{2}AB$ and $BD=\frac{1}{2}BC$ [$F$, $D$ are midpoints].

   1 mark

3. In $△FBD$ and $△DEF$: $FB=\frac{1}{2}AB=DE$, $BD=\frac{1}{2}BC=EF$, $DF=FD$ (common).
   So $\frac{FB}{DE}=\frac{BD}{EF}=\frac{DF}{FD}=1$. Hence $△FBD\sim △DEF$ [SSS similarity]. (In fact the two triangles are congruent.)

   1 mark

4. In $△DEF$ and $△ABC$: $\frac{DE}{AB}=\frac{1}{2}$, $\frac{EF}{BC}=\frac{1}{2}$, $\frac{FD}{CA}=\frac{1}{2}$.
   Since all three ratios are equal, $△DEF\sim △ABC$ [SSS similarity]. Hence proved.

   1 mark

Answer: $△FBD\sim △DEF$ and $△DEF\sim △ABC$ (SSS similarity, using the midpoint theorem).

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)
• Topic: Problems on Similar Triangles