Step-by-step explanation:
BC is the bisector of [tex] \angle ABQ[/tex]
[tex] \therefore \angle ABC \cong \angle CBQ..(1)\\
\angle CBQ \cong \angle CAB..(2) \\(\angle 's\: in \: alternate\: segments) \\[/tex]
From equations (1) & (2)
[tex] \angle ABC \cong \angle CAB\\
\therefore AC = BC [/tex]
(Because sides opposite to equal angles are equal)
Hence Proved
