147. The Triangle's Inner Circle · hugging all three sides perfectly

Try to break it

Drag A, B, or C. The three angle bisectors always meet at one point I, and the circle centred at I touches all three sides exactly once. Try to make the circle miss a side — you can't. Every triangle has exactly one incircle.

How you build it

Construct the incircle of a triangle.

• Mark point A, the first vertex of the triangle.
• Mark point B, the second vertex of the triangle.
• Mark point C, the third vertex of the triangle.
• Draw segment AB of the triangle.
• Draw segment BC of the triangle.
• Draw segment CA to complete triangle ABC.
• With centre B, draw an arc crossing BA (at D) and BC (at E).
• Mark D where the arc from B crosses side BA.
• Mark E where the arc from B crosses side BC.
• Open the compass wider than half of DE. Click D as centre and swing an arc inside angle B.
• Keep the same radius. Click E as centre and swing an arc. Where it crosses the arc from D is point F.
• Mark F where the arcs from D and E intersect — this is on the angle bisector of B.
• Draw a ray from B through F. This bisects angle B and will pass through the incentre I.
• With centre C, draw an arc crossing CB (at G) and CA (at H).
• Mark G where the arc from C crosses side CB.
• Mark H where the arc from C crosses side CA.
• Open the compass wider than half of GH. Click G as centre and swing an arc inside angle C.
• Keep the same radius. Click H as centre and swing an arc. Where it crosses the arc from G is point J.
• Mark J where the arcs from G and H intersect — this is on the angle bisector of C.
• Draw a ray from C through J. This bisects angle C and meets ray BF at the incentre I.
• Mark point I where the two angle bisectors meet. This is the incentre — equidistant from all three sides.
• Drop a perpendicular from I to side BC. The foot P gives the inradius IP = r.
• Draw the incircle with centre I and radius IP. It touches all three sides of the triangle.

The proof, step by step

Prove that the incircle touches all three sides of the triangle.

• The incenter I is the point where the angle bisectors of the triangle meet.
• Any point on an angle bisector is equidistant from the two sides forming that angle.
• Since I lies on both bisectors, it is equidistant from all three sides of the triangle.
• The perpendicular distance from I to any side is the inradius. A circle with this radius centered at I touches all three sides.

Worked example

In triangle ABC, the incenter I is 5 cm from side BC. What is the radius of the incircle?

The incenter is equidistant from all three sides. Therefore, the perpendicular distance from I to BC (5 cm) is exactly the inradius.

• 3 cm
• 5 cm — correct
• 7 cm
• 10 cm