CBSE Class 10 Mathematics Standard 2023: question 28(b)

Hints

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Step-by-step solution

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1. E lies on AD produced, and $AD\parallel BC$, so $AE\parallel BC$. In $△ALE$ and $△CLB$: $\angle LAE=\angle LCB$ (alternate angles) and $\angle ALE=\angle CLB$ (vertically opposite). So $△ALE\sim △CLB$ (AA).
   $\Rightarrow \frac{AL}{CL}=\frac{EL}{BL}$ ……(i)

   1 mark

2. Also $DC\parallel AB$. In $△CLM$ and $△ALB$: $\angle LCM=\angle LAB$ (alternate angles) and $\angle CLM=\angle ALB$ (vertically opposite). So $△CLM\sim △ALB$ (AA).
   $\Rightarrow \frac{AL}{CL}=\frac{AB}{CM}$

   1 mark

3. Opposite sides of a parallelogram are equal, so $AB=CD$: $\frac{AL}{CL}=\frac{CD}{CM}$ ……(ii)

   ½ mark

4. From (i) and (ii): $\frac{EL}{BL}=\frac{CD}{CM}=\frac{2CM}{CM}=2$ [M is the midpoint of CD]. $\Rightarrow EL=2BL$. Hence proved.

   ½ mark

Answer: Using ΔALE ~ ΔCLB and ΔCLM ~ ΔALB with AB = CD = 2CM, EL = 2BL (proved).

Revise the chapter

• Chapter: Triangles (notes, formulas, where marks are lost and every past question)
• Topic: Basic Proportionality Theorem (Thales Theorem)