Steps of construction:
- Define a boundary portion \[AB\text{ }=\text{ }5\text{ }cm.\] Develop a right point \[SAB\]at point \[A.\]
- Draw a circular segment of span \[12\text{ }cm\]with \[B\]as its middle to cross\[SA\text{ }at\text{ }C\].
- Join \[BC\]to get \[ABC.\]
- Draw a beam \[AX\]making an intense point with \[AB,\]inverse to vertex \[C.\]
- Find \[3\] focuses,\[{{A}_{1}},\text{ }{{A}_{2}},\text{ }{{A}_{3}}\]
on line portion \[AX\]with the end goal that \[A{{A}_{1}}~=\text{ }{{A}_{1}}{{A}_{2}}~=\text{ }{{A}_{2}}{{A}_{3}}.\]
- Join\[{{A}_{3}}B.\]
- Define a boundary through\[{{A}_{2}}\]
corresponding to\[{{A}_{3}}B\] meeting \[AB\text{ }at\text{ }B’.\]
- Through\[B’\], define a boundary corresponding to \[BC\]meeting \[AC\text{ }at\text{ }C’.\]
- Triangle \[AB’C’\]is the necessary triangle.