STEPS OF CONSTRUCTION
(i) Draw a line XY.
(ii) Mark any point D on XY.
(iii) From D, draw `DE bot XY`.
(iv) From D, set off `DA=4` cm, cutting DE at A.
(v) Construct `angleDAB=30^(@)` and `angleDAC=30^(@)` meeting XY at B and C respectively.
Then, `Delta ABC` is the required equilateral triangle.
Verification :
On measuring, we find that
`angleA=angleB=angleC=60^(@)`
and `AB=BC=CA=4.5` cm
Justification:
In `Delta DAB`, we have
`angleABD+angleBDA+angleDAB=180^(@) implies angleABD+90^(@)+30^(@)=180^(@)`
`implies angleABD=180^(@)-120^(@)=60^(@)`
In `Delta DAC`, we have
`angleACD+angleCDA+angleDAC=180^(@) implies angleACD+90^(@)+30^(@)=180^(@)`
`implies angleACD=180^(@)-120^(@)=60^(@)`.
In `Delta ABC`, we have `angleA=angleB=angleC=60^(@)`.
Hence, `Delta ABC` is an equilateral triangle.
