Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.

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Solution

Steps of construction
1. Draw a line segment AB = 7 cm
2. Draw a ray AX, making acute ∠BAX
3. Along AX mark 3 + 5 = 8 Points
$A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}$ such that
$A A_{1} = A A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5} = A_{5} A_{6} = A_{6} A_{7} = A_{7} A_{8}$
4. Join $A_{8} B$
5. From $A_{3}$ . Draw $A_{3} C | | A_{8} B$ meeting AB at C.
[By making an angle equal to $∠ B A_{8} A$ at $A_{3}$ ]
Then C is the point on AB which divides it in the ratio 3:5.
Thus AC : BC = 3 : 5

Justification
Let $A A_{1} = A_{1} A_{2} = A_{2} A_{4} = A_{7} A_{8} = x$
In $Δ A B A_{8}$ , we have $A_{3} C | | A_{8} B$
$∴ \frac{A C}{C B} = \frac{A A_{3}}{A_{3} A_{8}} = \frac{3 X}{5 X} = \frac{3}{5}$
Hence AC : CB = 3 : 5

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