Construct a triangle ABC with side …

 
To construct: To construct a triangle ABC with side BC = 7 cm, {tex}\angle B = 45^\circ {/tex} and {tex}\angle A = 105^\circ {/tex} and then a triangle similar to it whose sides are {tex}\frac{4}{3}{/tex} of the corresponding sides of the first triangle ABC.
Steps of construction:

1. Draw a triangle ABC with side BC = 7 cm, {tex}\angle B = 45^\circ {/tex},  {tex}\angle A= 105^\circ {/tex}and  {tex}\angle C= 30^\circ {/tex}
2. From any ray BX, making an acute angle with BC on the side opposite to the vertex A.
3. Locate 4 points B₁, B₂, B₃ and B₄ on BX such that BB₁ = B₁ B₂ = B₂ B₃ = B₃B.
4. Join B₃ C and draw a line through the point B₄, 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_4}C'||{B_3}C{/tex} [By construction]
   {tex}\therefore \triangle B{B_4}C' \sim \triangle B{B_3}C{/tex} [AA similarity]
   {tex}\therefore \frac{{B{B_4}}}{{B{B_3}}} = \frac{{BC'}}{{BC}}{/tex} [By Basic Proportionality Theorem]
   But {tex}\frac{{B{B_4}}}{{B{B_3}}} = \frac{4}{3}{/tex} [By construction]
   Therefore, {tex}\frac{{BC'}}{{BC}} = \frac{4}{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{{A'B}}{{AB}} = \frac{{A'C'}}{{AC}} = \frac{{BC'}}{{BC}} = \frac{4}{3}{/tex} [From eq. (i)]