CBSE Class 10 Mathematics Standard 2023: question 35(b)

Hints

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Step-by-step solution

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1. Figure: Draw parallelogram ABCD with M the mid-point of CD. Join BM and extend it to meet AD produced at E. Draw diagonal AC, which cuts BE at L.

   1 mark

2. In $△BMC$ and $△EMD$:
   $MC=MD$ [M is the mid-point of CD]
   $\angle CMB=\angle DME$ [vertically opposite angles]
   $\angle MBC=\angle MED$ [alternate angles, since $BC\parallel AE$]
   So $△BMC\cong △EMD$ [AAS], and therefore $BC=DE$ [CPCT].

   1 mark

3. Opposite sides of a parallelogram are equal, so $AD=BC$. Hence $AD=DE$, which gives $AE=AD+DE=2BC$.

   1 mark

4. In $△AEL$ and $△CBL$:
   $\angle LAE=\angle LCB$ [alternate angles, $AE\parallel BC$]
   $\angle ALE=\angle CLB$ [vertically opposite angles]
   So $△AEL\sim △CBL$ [AA similarity].

   ½ mark

5. Corresponding sides are proportional: $\frac{EL}{BL}=\frac{AE}{CB}=\frac{2BC}{BC}$.

   ½ mark

6. So $\frac{EL}{BL}=2$, i.e. $EL=2BL$. Hence proved.

   1 mark

Answer: $△BMC\cong △EMD$ gives $AE=2BC$; then $△AEL\sim △CBL$ gives $\frac{EL}{BL}=2$, i.e. $EL=2BL$.

Revise the chapter

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