CBSE Class 10 Mathematics Standard 2026: question 33

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Step-by-step solution

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1. Given: In $\Delta ABC$, AD is a median, so $BD=DC$. X lies on AD with $AX:XD=2:3$. BX produced meets AC at Y.

   To prove: $BX=4XY$.

2. Construction: Through D, draw $DZ\parallel BY$ meeting AC at Z.

   ½ mark

3. In $\Delta CBY$, $DZ\parallel BY$. By the Basic Proportionality Theorem,
   $$\frac{CD}{DB}=\frac{CZ}{ZY}$$
   Since D is the midpoint of BC, $\frac{CD}{DB}=1$, so $\frac{CZ}{ZY}=1$ and Z is the midpoint of CY.

   1 mark

4. In $\Delta CBY$, D and Z are the midpoints of CB and CY. By the midpoint theorem, $DZ=\frac{1}{2}BY$, that is, $BY=2DZ$ ...(i)

   ½ mark

5. In $\Delta ADZ$, $XY\parallel DZ$ (XY is part of BY).

   So $\angle AXY=\angle ADZ$ and $\angle AYX=\angle AZD$ [corresponding angles], with $\angle A$ common.

   Hence $\Delta AXY\sim \Delta ADZ$ [AA similarity], which gives
   $$\frac{AX}{AD}=\frac{XY}{DZ}$$

   1 mark

6. Since $AX:XD=2:3$, we have $AD=2+3=5$ parts, so $\frac{AX}{AD}=\frac{2}{5}$.

   Therefore $\frac{2}{5}=\frac{XY}{DZ}$, which gives $DZ=\frac{5}{2}XY$ ...(ii)

   ½ mark

7. Substituting (ii) in (i):
   $$BY=2\times \frac{5}{2}XY=5XY$$

   1 mark

8. Now $BX=BY-XY=5XY-XY=4XY$.

   Hence proved.

   ½ mark

Answer: Proved: BX = 4 XY (using DZ ∥ BY, BY = 2DZ and DZ = (5/2)XY).

Revise the chapter

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