In any triangle the three angle bisectors meet at a point (the incentre), and a circle from this point will just touch the three sides of the triangle tangentially.

In triangle ABC, construct the bisectors of angle B and angle C. Call the point where they cross, O.

Construct a perpendicular OS, from O to the side AB.

On BC mark the point Q so that BS = BQ.

Consider the triangles BOS and BOQ.

BO is a common side, angle SBO = angle QBO ( bisection) and BS = BQ by construction.

So the triangles are congruent (SAS).

So angle BQO = BSO = 90. And SO = QO.

So a circle centred on O will go through S and Q, and BQ is a tangent to this circle. (BS was constructed to be a tangent.)

Construct R on AC so that CR = CQ.

By a similar argument, triangles CQO and CRO are congruent, so RO = QO. Therefore, the circle also touches AC tangentially at R.

Consider triangles ASO and ARO.

AO is common. SO = RO. angle ARO = angle ASO = 90. So the triangles are congruent (SSR).

Therefore SAO = RAO, so AO is the bisector of the angle BAC, Therefore the bisectors are concurrent.

Bob

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