Geometry Proof about Circles

Let the angle bisector be AD, where D intersects the circle

Join DB  and DC 

And we have two triangles - BAD  and CAD

Then

AB = AC

And angle BAD  = angle CAD

And AD is common

So...by SAS, triangle BAD is congruent to triangle CAD

But  angle DBA  = angle DCA...so....they intersect equal arcs

So....arc DBA  = arc DCA

But the sum of these arcs = 360°

So.....each angle intercepts an arc = 180°

But......the measures of these angles = 1/2 of their intercepted arcs = 90°

But...in a circle....an inscribed angle of 90° has its enpoints on a diameter

So...DA  must be a diameter

And a diameter always passes through the center of a circle

But DA is the angle bisector

So....DA passes through the center of the circle