$I(1,0)$ is the centre of circle of triangle $ABC$, the equation of $BI$ is $x - 1 = 0$ and equation of $CI$ is $x - y - 1 = 0$, then angle $BAC$ is:
Step-by-Step Solution
Key Concept: The angle at the incenter relates to the angle at the opposite vertex via $\angle BIC = \frac{\pi}{2} + \frac{A}{2}$.
Given $\angle BIC = \frac{3\pi}{4} = \frac{\pi}{2} + \frac{A}{2}$, we solve to find $A = \frac{\pi}{2}$, making $\angle BIC = \frac{\pi}{2} + \frac{A}{2}$. From the diagram, point $I(1,0)$ is the incenter, and the angle calculation yields $\angle BAC = \frac{\pi}{2}$. Therefore, $\angle ABC = \frac{\pi}{2}$ (right angle at $B$), and the configuration confirms the geometric relationship.
Correct Answer: 3