We assume that points P and Q are marked on side BC, and point O on bisector AP or its extension.
Extend BO crossing side AC or its extension at point D.
In ΔABD, ∠BAO=∠DAO and AO⊥BD So, ΔABO≡ΔADO (ASA) Thus, AB=AD, BO=OD
In ΔBCD, OQ∥DC, so BO : OD=BQ : QC, and BO=OD
Therefore, BQ=QC Q.E.D.