AB, AC are two equal chords of a circle. Prove that the line segment which bisects ∠BAC passes through the center of a circle.

FreezingTNT Nov 16, 2018

#1**+2 **

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

CPhill Nov 16, 2018