Let the angle-bisectors of angles C and B meet at I and join AI. We need
to show that AI is the angle bisector of angle A.
From I draw perpendiculars to the three sides of the triangle to meet the
sides at P, Q and R.
By our construction of the perpendiculars APIQ, PIRC and BRIQ are all
cyclic quadrilaterals.
Hence angle IRP = angle ICP (=x) and angle IPR = angle ICR (=x) and so IP=IR.
Similarly IR = IQ (using the cylcic quad BRIQ). Hence IP = IQ.
In cyclic quad APIQ: angle QAI = angle QPI, and angle IQP = angle IAP, BUT
angle QPI = angle IQP (IP = IQ) and so angle QAI = angle IAP -- AI is the
angle bisector of angle A!
Furthemore it is clear that I is equidistant from the three sides of the
triangle and hence the center of the inscribed circle.