343. The Incenter's Promise · equidistant from every side

Try to break it

Drag A, B, or C. The three angle bisectors always meet at one point I, and the perpendicular distance from I to each side is the same — that distance is the inradius. Try to drag the vertices so the inscribed circle pokes through or pulls away from a side; impossible. I sits at equal perpendicular distance from all three sides.

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 A, draw an arc crossing AB (at P) and AC (at Q).
• Mark P where the arc from A crosses side AB.
• Mark Q where the arc from A crosses side AC.
• Open the compass wider than half of PQ. Click P as centre and swing an arc inside angle A.
• Keep the same radius. Click Q as centre and swing an arc. Where it crosses the arc from P is point R.
• Mark R where the arcs from P and Q intersect — this is on the angle bisector of A.
• Draw a ray from A through R. This bisects angle A and will pass through the incentre I.
• With centre B, draw an arc crossing BA (at S) and BC (at T).
• Mark S where the arc from B crosses side BA.
• Mark T where the arc from B crosses side BC.
• Open the compass wider than half of ST. Click S as centre and swing an arc inside angle B.
• Keep the same radius. Click T as centre and swing an arc. Where it crosses the arc from S is point U.
• Mark U where the arcs from S and T intersect — this is on the angle bisector of B.
• Draw a ray from B through U. This bisects angle B and meets ray AR at the incentre I.
• Mark point I where the two angle bisectors meet. This is the incentre — equidistant from all three sides.
• From I, drop a perpendicular to side BC. The foot D gives the inradius ID = r.
• Draw the incircle with centre I and radius ID. It touches all three sides of the triangle.

The proof, step by step

Prove that the constructed circle touches all three sides of the triangle.

• The incenter I lies on the bisector of ∠A, so it is equidistant from sides AB and AC.
• Similarly, I lies on the bisector of ∠B, so it is equidistant from sides BA and BC.
• Therefore, I is equidistant from all three sides. This common distance is the inradius.
• A circle centered at I with this radius touches all three sides, forming the inscribed circle.

Worked example

In △ABC, the angle bisectors of ∠A and ∠B intersect at I. If the perpendicular distance from I to side BC is 5.4 cm, what is the distance from I to side AB?

The incenter is equidistant from all sides of the triangle. Since the distance to BC is 5.4 cm, the distance to AB must also be exactly 5.4 cm.

• 2.7 cm
• 5.4 cm — correct
• 8.1 cm
• 10.8 cm