Draw a line segment of length …

We have to draw a line segment of length 7 cm.Then,we have to find  a point P on it, which divides it in the ratio 3 : 5.
Steps of construction:

1. Draw a line segment {tex}AB=7{/tex} cm.
2. Draw a ray AX, making an acute {tex}\angle BAX{/tex} with AB.
3. Mark {tex}3+5=8{/tex} points, i.e, {tex}A_1, A_2, A_3, A_4 ...A_8{/tex} on AX, such that {tex}AA_1 = A_1A_2 = A_2A_3 = A_3A_4 ... = A_7A_8{/tex}
4. Join A{tex}_8{/tex}B
5. From A₃, draw {tex}A_3P || A_8B{/tex} which intersects AB at point P [ by making an angle at A₃ equal to {tex}\angle AA_8B{/tex} 
   Then, P is the point on AB which divides it in the ratio 3:5. So, {tex}AP : PB = 3:5{/tex}

Justification: In {tex}\triangle ABA_8{/tex}, we have {tex}A_3P || A_8B{/tex} 
{tex}\therefore \frac{AP}{PB}=\frac{AA_3}{A_3A_8}{/tex} [ by basic proportionality theorem]
By construction, {tex}\frac{AA_3}{A_3A_8}=\frac{3}{5}{/tex} 
Hence, {tex}\frac{AP}{PB}=\frac{3}{5}{/tex}