Construct a triangle sides 4,5,6cm and …

To construct: To construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are {tex}\frac{2}{3}{/tex} of the corresponding sides of the first triangle.
 
Steps of construction :

1. Draw a triangle ABC of sides 4 cm, 5 cm and 6 cm.
2. From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
3. Locate 3 points B₁, B₂ and B₃ on BX such that BB₁ = B₁ B₂ = B₂ B₃ .
4. Join B₃C and draw a line through the point B₂, draw a line parallel to B₃ C intersecting BC at the point C'.
5. Draw a line through C' parallel to the line CA to intersect BA at A'.
   Then, A'BC' is the required triangle.

Justification :
{tex}\because {B_3}C||{B_2}C'{/tex} [By construction]
{tex}\therefore \frac{{B{B_2}}}{{{B_2}{B_3}}} = \frac{{BC'}}{{C'C}}{/tex} [By Basic Proportionality Theorem]
But {tex} \frac{{B{B_2}}}{{{B_2}{B_3}}} = \frac{2}{1}{/tex} [By construction]
Therefore, {tex}\frac{{BC'}}{{C'C}} = \frac{2}{1} \Rightarrow \frac{{C'C}}{{BC'}} = \frac{1}{2} \Rightarrow \frac{{C'C}}{{BC'}} + 1 = \frac{1}{2} + 1{/tex}
{tex} \Rightarrow \frac{{C'C + BC'}}{{BC'}} = \frac{{1 + 2}}{2} \Rightarrow \frac{{BC}}{{BC'}} = \frac{3}{2} \Rightarrow \frac{{BC'}}{{BC}} = \frac{2}{3}{/tex}}...}...} ......(i)
{tex}\because CA||C'A'{/tex} [By construction]
{tex}\therefore \triangle BC'A' \sim \triangle BCA{/tex} [AA similarity]
{tex}\therefore \frac{{AB'}}{{AB}} = \frac{{A'C'}}{{AC}} = \frac{{BC'}}{{BC}} = \frac{2}{3}{/tex} [From eq. (i)]