Draw a triangle ABC in which BC= 9 cm , Angle B = 60 and AB = 6 cm. Then construct another triangle whise sides are 2/3 of corresponding sides of Triangle ABC.

Answer  :

We follow these steps to construct angle ABC   .

Step 1 :  Draw a line BC = 9 cm  .

Step 2 :  Now take any radius ( Less than half of BC ) and take center  " B "  draw an arc that intersect our line BC at  " P " , With same radius and center is " P "  draw another arc that intersect our previous  arc at "  Q " .

Step 3 :    Join BQ and extend it to D

Step 4 :  Take radius =  6 cm and center "  B "  we draw an arcs that intersect our line BD at "  A "  Now we join AC , and get triangle ABC .

Now We construct another triangle with scale factor $\frac{2}{3}$ , As :
              
Step 5 :Below BC,  draw a line BX  As  $\angle$ CBX is any acute angle and draw three arcs of equal radii on line BX that intersect at X₁ . X₂ and  X₃ AS :

  BX₁  = X₁ X₂ = X₂ X₃ 

Step 6 : Join X₃ to C and than draw a line from X₂  as parallel to X₃C that intersect our line BC at C' .

Step 7 : Draw line from C' as parallel to CA that intersect line AB at A'  .



We get triangle A'BC' whose sides are $\frac{2}{3}$ of corresponding sides of triangle ABC .