The Inscribed Circle.
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.
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