# To divide a line segment AB in the ratio $5:6$, draw a ray AX such that $\angle BAX$ is an acute angle, then draw a ray BY parallel to AX and the points $A_{1},A_{2},A_{3}\cdots$and $B_{1},B_{2},B_{3}\cdots$are located at equal distances on ray AX and BY, respectively. Then the points joined are (A) A5 and B6            (B) A­ and B5               (C) A4 and B5             (D) A5 and B4

Solution

Given: $\angle BAX$ and $\angle ABY$ both are acute angles and AX parallel to BY
The required ratio is $5:6$
Let m = 5 , n = 6

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray $BY\parallel AX$.
3. Locate the points $A_{1},A_{2},A_{3},A_{4},A_{5}$ on AX at equal distances
4. Locate the points $B_{1},B_{2},B_{3},B_{4},B_{5}$ on BY at distance equal to the distance between points on AX line.
5. Join $A_{5}B_{6}$.
Let it intersect AB at a point C in figure.
Then $AC:CB= 5:6$
Here $\bigtriangleup AA_{5}C$  is similar to $\bigtriangleup BB_{6}C$
Then $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}= \frac{AC}{BC}$
$\therefore$by construction $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}$
$\therefore \frac{AC}{BC}= \frac{5}{6}$
$\therefore$ Points joined one A5 and B6.

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