CBSE Class 10 Mathematics Basic 2024: question 34(b)

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Step-by-step solution

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1. Figure: Draw $△ABC$ with CD bisecting $\angle ACB$ (D on AB). Mark $\angle ACD=\angle 1$ and $\angle DCB=\angle 2$. Draw $△FEG$ with GH bisecting $\angle EGF$ (H on FE). Mark $\angle FGH=\angle 3$ and $\angle HGE=\angle 4$.

   Given: $△ABC\sim △FEG$, and CD, GH are the bisectors of $\angle ACB$ and $\angle EGF$.

   1 mark

2. (i) Since $△ABC\sim △FEG$, corresponding angles are equal, so $\angle ACB=\angle FGE$.

   ½ mark

3. Halving both angles gives $\frac{1}{2}\angle ACB=\frac{1}{2}\angle FGE$ [CD and GH are bisectors].

   Hence $\angle 1=\angle 3$ and $\angle 2=\angle 4$.

   1 mark

4. In $△ACD$ and $△FGH$:

   • $\angle A=\angle F$ [corresponding angles of similar triangles]
   • $\angle 1=\angle 3$ [proved above]

   So $△ACD\sim △FGH$ [AA similarity criterion].

   1 mark

5. Corresponding sides of similar triangles are proportional, so $\frac{CD}{GH}=\frac{AC}{FG}$.

   ½ mark

6. (ii) In $△DCB$ and $△HGE$:

   • $\angle B=\angle E$ [corresponding angles, as $△ABC\sim △FEG$]
   • $\angle 2=\angle 4$ [proved above]

   So $△DCB\sim △HGE$ [AA similarity criterion].

   1 mark

Answer: (i) $△ACD\sim △FGH$ (AA), so $\frac{CD}{GH}=\frac{AC}{FG}$; (ii) $△DCB\sim △HGE$ (AA).

Revise the chapter

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