Need help with Euclidean geometry Here's the problem:

Consider triangle ABC inscribed in circle C(O, r). The angle bisector of ∠A (resp. ∠B,
∠C) intersects the circle C(O, r) in the points A and A' (resp. B and B', C and C'). Prove that m(∠A'B'C') = 90 − 0.5m(∠ABC).

I wasn't given a diagram, but I drew a rough sketch of what I think it looks like.

I don't know where to start with this. Any advice would be appreciated. This is for a college-level course, by the way.
 A: We have: $m(\angle B) = \dfrac{\text{arc}(AC)}{2} \implies m(\angle B') = \dfrac{\text{arc}(A'C')}{2}=\dfrac{\text{arc}(BC')+\text{arc}{BA'}}{2}= \dfrac{1}{2}\cdot \left(\dfrac{\text{arc}{BA}}{2}+\dfrac{\text{arc}{BC}}{2}\right)= \dfrac{1}{4}\cdot \left(360^{\circ} - \text{arc}{AC}\right)= 90^{\circ} - \dfrac{\text{arc}{AC}}{4}=90^{\circ} - \dfrac{m(\angle B)}{2}$.
A: 
We can make repeated use of circle-chord theorems.  If we draw in chord $ \ \overline{A'C'} \ \ , $ we find that $ \ \angle A'BC' \ = \ 180º - \angle A'B'C' \ \ . $  We wish to find the measure of
$$ \angle ABC \ \ = \ \ \angle A'BC' \ - \ \angle ABC' \ - \ \angle A'BC \ \ . $$
We can also say that $ \ \angle AC'B \ = \ 180º - \angle ACB \ $ and $ \ \angle BA'C \ = \ 180º - \angle BAC \ \ .  $  Since $ \ A' \ $ and $ \ C' \ $ bisect the arcs $ \ \widehat{AC'B} \ $ and $ \ \widehat{BA'C} \ \ , $ the triangles $ \ \triangle AC'B \ $ and $ \ \triangle BA'C \ $ are isosceles, from which it follows that $ \ \angle ABC' \ = \ \frac12 · (180º - \angle AC'B) \ $ and $ \ \angle A'BC \ = \ \frac12 · (180º - \angle BA'C) \ \ . $
Using these results in the above equation produces
$$ \angle ABC \ \ = \ \ (180º - \angle A'B'C') \ - \ \frac12 · (180º - \angle AC'B) \ - \ \frac12 · (180º - \angle BA'C) $$
$$  = \ \ \frac12 · (\angle AC'B + \angle BA'C) \ - \angle A'B'C'  $$
for which the circle-chord theorem gave us
$$  = \ \ \frac12 · ( \ [180º - \angle ACB] + [180º - \angle BAC] \ ) \ - \angle A'B'C' $$ $$ = \ \ 180º \ - \ \frac12 · ( \ \angle ACB +  \angle BAC \ ) \ - \angle A'B'C' \ \ . $$
But the sum of the angles in parentheses is the supplement of $ \ \angle ABC \ \ , $ so
$$ \angle ABC \ \ = \ \ 180º \ - \ \frac12 ·  ( \ 180º \ - \  \angle ABC  \ ) \ - \angle A'B'C' \ \ = \ \ 90º \ + \ \frac12 ·  \angle ABC  \  -  \ \angle A'B'C'  $$
$$ \Rightarrow \ \ \angle A'B'C' \ \ = \ \ 90º \ - \ \frac12 ·  \angle ABC  \ \ . $$
ADDENDUM (5/7) --

We can also approach this using a different circle-chord theorem.  From the center of the circle $ \ O \ \ , $ draw radii to the points $ \ A , B , C , A' , B' , C' \ \ . $  We will designate the "direction" of the radius  $ \ \overline{OA} \ $ as "zero".
If we label the measures of the angles in $ \ \triangle ABC \ $ as $  \ m(\angle BAC) \ = \ \alpha \ \ , \ \ m(\angle ABC) \ = \ \beta \ \ , $ $ m(\angle BCA) \ = \ \gamma \ \ , \ \ $ then $  \ m(\angle BOC) \ = \ 2\alpha \ \ , \ \ m(\angle AOC) \ = \ 2\beta \ \ , $ $ m(\angle BOA) \ = \ 2\gamma \ \ . \   $   Measuring clockwise, the direction of radius $ \ \overline{OB} \ $ is then $ \ 2\gamma \ $ and that of  $ \ \overline{OC} \ $ is $ \ 2\gamma \ + \ 2\alpha \ \ . $   (Note that if we continue in this manner, the direction of $ \ \overline{OA} \ $ is $ \ 2\gamma \ + \ 2\alpha \ + \ 2\beta $ $ = \ 2·(\alpha  +  \beta  +  \gamma) \ = \ 360º \ , $  as it should be.)
As has been said, the points $ \ A' , B' , C' \ $ bisect the respective arcs they lie in.  Hence, the direction of $ \ \overline{OC'} \ $ is $ \ \gamma \  , $ that of $ \ \overline{OA'} \ $ is $ \ 2\gamma \ + \ \alpha \ , $ and that of $ \ \overline{OB'} \ $ is $ \ 2\gamma \ + \ 2\alpha \ + \ \beta \ . $
We may then calculate that $ \ m(\angle A'B'C') \ $ is half of the difference between the directions of $ \ \overline{OA'} \ $ and $ \ \overline{OC'} \ \ , $ which is
$$   m(\angle A'B'C') \ \ = \ \ \frac12·m(\angle A'OC') \ \ = \ \ \frac12·( \ [2\gamma \ + \ \alpha] \ - \ \gamma \ ) \ \ = \ \ \frac12·(   \alpha    +   \gamma  ) \ \ . $$
Since $ \ \alpha  +  \beta  +  \gamma \ = \ 180º \ \ , $ we have $  \ m(\angle A'B'C') \   = \  \frac12·( \  180º - \beta \  ) \ \ ; $ referring back to our angle designations, this gives us
$$  m(\angle A'B'C') \ \  = \ \ 90º \ - \ \frac12·m(\angle ABC) \ \ . $$
(Note that this is consistent with the directions of the radii to $ \ A' , B' , C' \ \ : \ $ by similar reasoning to the above, we have
$  m(\angle B'A'C') \    =   \ 90º \ - \ \frac12·m(\angle BAC)  \ = \ 90º \ - \ \frac12·\alpha \ ; $ then  $ \ m(\angle B'OC') \     = \ 180º \ - \ \alpha \ $ and continuing clockwise from the direction of
$ \ \overline{OB'} \ $ gives the direction of $ \ \overline{OC'} \ $ as
$$ ( \ 2\gamma \ + \ 2\alpha \ + \ \beta \ ) \ + \ ( \ 180º \ - \ \alpha \ ) \ \ = \ \ 180º \ + \ 2\gamma \ + \  \alpha \ + \ \beta \    $$
$$ = \ \ 180º \ + \ ( \ \alpha \ + \ \beta \ + \ \gamma \ ) \ + \ \gamma \ \ = \ \ 360º \ + \ \gamma \ \ ,  $$
again as it should be. )
[This second solution is more along the lines of the answers posted by Wang YeFei and Wizard0001 . ]
A: $$\angle A'B'C'=\angle A'B'B+\angle BB'C'=\angle A'AB+\angle BCC'=\frac{\angle CAB}{2}+\frac{ACB}{2}=\frac{180^{\circ}-\angle ABC}{2}=90^{\circ}-\frac{\angle ABC}{2}$$
