Draw triangle ABC
Draw angle bisectors AD, CE which cross at O
From O draw perpendiculars OP to AB and ON to AC
Consider ∇ANO and ∇APO
<NAO = <EAO ..........Construct
<ANO = 1 right angle ...Construct
<APO = 1 right angle ...Construct
∴ <AON = <AOP ...(sum of angles of triangle = 180º)
AO = OA .............Common
∴ ∇ANO = ∇APO are congruent (2 ang.s & inc. side equal)
∴ NO = PO
From O draw perpendicular OM to BC
Using similar argument to above
∇CNO = ∇CMO
PO = NO (Proven above)
∴ PO = MO = NO
Draw BO
PO = MO ......(proven above)
∴ ∇OMP is an isosceles triangle
∴ <OMP = <OPM
∴ <BMP = <BPM as <BMO = <BPO = 1 rt. ang.
∴ ∇BMP is an isosceles triangle
∴ BM = BP
∴ ∇BMO = ∇BPO ........ (2 sides and incl. angle equal)
∴ <MBO = <PBO
∴ BO is the bisector of <ABC
QED
Taken from Wentworth