Theory:

Incentre: It is the point of concurrency of a triangle formed by the intersection of its three angle bisectors of a triangle. It is denoted by \(I\).
Incircle:It is the circle drawn inside the triangle with incentre as its centre.
Working rule to construct the Incircle of a Triangle
Case 1:
 
Given the length of two sides of a triangle and the measure of one of its interior angle.
Example:
Construct the incircle of a triangle \(ABC\) given \(AB = 8\) \(cm\), \(BC = 6\) \(cm\) and \(\angle BAC = 55^{\circ}\). Locate its incentre and find the radius.
 
Construction:
 
Step 1: Draw a rough figure for the given question to get the picture of the triangle that is to be constructed.
 
Step 2: Draw a line segment \(AB = 8\) \(cm\) using the ruler.
 
Step 3: With \(A\) as centre, mark an angle \(55^{\circ}\) using the protractor and name it as \(X\). Now join \(AX\).
 
Step 4: With \(B\) as centre, measure \(6\) \(cm\) in the compass and cut an arc intersecting \(AX\) and mark it as \(C\).
 
Step 5: Draw the angle bisectors of any two angles of the triangle \(A\) and \(B\) (say).
 
Step 6: Mark the intersecting point of the angle bisectors as \(I\), which is the incentre of the given triangle.
 
Step 7: Draw a perpendicular bisector from \(I\) to one of the sides of the triangle \(AB\) (say) to meet \(AB\) at \(D\).
 
Step 8: Measure the length of \(ID\) to get the inradius of the incircle.
 
Step 9: With \(I\) as centre and \(ID\) as radius, draw the incircle of the triangle.
 
Incentre Case i GIF.gif
Case 2:
 
Given the length of a side of a triangle and the measure of two of its interior angle.
Example:
Construct the incircle of a triangle \(ABC\) given \(AB = 6\) \(cm\), \(\angle BAC = 55^{\circ}\) and \(\angle CBA = 60^{\circ}\). Locate its incentre and find the radius.
 
Construction:
 
Step 1: Draw a rough figure for the given question to get the picture of the triangle that is to be constructed.
 
Step 2: Draw a line segment \(AB = 6\) \(cm\) using the ruler.
 
Step 3: With \(A\) as centre, mark an angle \(55^{\circ}\) using the protractor and name it as \(X\). Now join \(AX\).
 
Step 4: With \(B\) as centre, mark an angle \(60^{\circ}\) using the protractor and name it as \(Y\). Now join \(BY\).
 
Step 5: Mark the intersecting point of \(AX\) and \(BY\) as \(C\).
 
Step 6: Draw the angle bisectors of any two angles of the triangle \(A\) and \(B\) (say).
 
Step 7: Mark the intersecting point of the angle bisectors as \(I\), which is the incentre of the given triangle.
 
Step 8: Draw a perpendicular bisector from \(I\) to one of the sides of the triangle \(AB\) (say) to meet \(AB\) at \(D\).
 
Step 9: Measure the length of \(ID\) to get the inradius of the incircle.
 
Step 10: With \(I\) as centre and \(ID\) as radius draw the incircle of the triangle.
 
Incenter case ii gif.gif