The radius of the circle passing through the incentre $I$ of $\triangle ABC$ and through the end points of $BC$ is given by:
Step-by-Step Solution
Key Concept: The inscribed angle theorem relates the central angle to the inscribed angle, enabling application of the Sine Rule to find the circumradius.
Let $O$ be the circle's center with radius $R_1$ and $BC = a$. Since $\angle BOC = 2\angle BDC = (180° - A)$, we get $\angle BDC = 90° - \frac{A}{2}$. Applying the Sine Rule in $\triangle BOC$: $\frac{a}{\sin\angle BDC} = 2R_1$, which yields $R_1 = \frac{a}{2}\sec\frac{A}{2}$.
Correct Answer: 2