Originally Posted by
earboth Hello,
I assume that CE is a tangent to the circle.
If so $\displaystyle \bigtriangleup (ABC)$ is an isosceles triangle with base $\displaystyle AB$.
Let M be the centre of the circle and CE is tangent to the circle:
1. $\displaystyle (CM) \bot (CE)$
2. $\displaystyle (CE) \parallel (AB)$
3. $\displaystyle (CM) \bot (AB)$
4. (CM) is perpendicular bisector of (AB) thus $\displaystyle \bigtriangleup (ABC)$ is an isosceles triangle.
5. thus $\displaystyle \angle(CBA) = \angle(BAC)$
6. $\displaystyle \angle(DCA)$ is an exterior angle.
7. $\displaystyle \angle(CBA) + \angle(BAC) = \angle(DCA)$
8. Therefore: $\displaystyle \angle(CBA) = \angle(BAC) = 55^\circ$
9. Therefore $\displaystyle \angle(BMC) = 2 \cdot \angle(BAC) = 110^\circ$
10.The arc(BC) belongs to the angle $\displaystyle \angle(BMC)$. You only have to convert the measure in degree into radians:
$\displaystyle arc(BC) = \frac{110^\circ}{360^\circ} \cdot 2 \pi = \frac{11}{18} \pi \approx 1.92$
Maybe you should seek for the tangent-chord-theorem(?) [I don't know if this is the appropriate name in English] which states that the angle between the chord (here BC, name of angle $\displaystyle \angle(BCF)$) is as large as the angle at the circumference (here $\displaystyle \angle(BAC)$).