Mathematics — CBSE Class 10 board question

Since XZ ∥ BC, by Basic Proportionality Theorem (Thales Theorem) in △ABC:

$$\frac{AX}{XB} = \frac{AZ}{ZC} = \frac{3}{2}$$

In △ABM, since XY ∥ BM (as XZ ∥ BC), by BPT:

$$\frac{XY}{BM} = \frac{AX}{AB} = \frac{AZ}{AC} = \frac{3}{3+2} = \frac{3}{5}$$

$$XY = \frac{3}{5} \times BM = \frac{3}{5} \times 3 = \frac{9}{5} = \mathbf{1.8 \text{ cm}}$$

Source: Chapter 6, Section 6.3 (Basic Proportionality Theorem)

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• Since XZ ∥ BC, point Y (where AM meets XZ) lies on a line parallel to BC inside △ABM. So XY ∥ BM.
• Apply BPT in △ABC to get AX/AB = AZ/AC = 3/5.
• Then in △ABM, XY ∥ BM gives XY/BM = AX/AB = 3/5.
• Multiply by BM = 3 cm to get XY = 1.8 cm.
• Key: always identify the correct triangle before applying BPT.